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CAT Online Test Series 4  Free CAT Quiz Series 4  CAT Free Mock Test. Common Admission Test (CAT) is one of the most challenging and competitive MBA entrance test in our country. Check your level of preparation for CAT with free All India Mock test 2020. CAT Exam Online Test 2020, CAT Free Mock Test Exam 2020. CAT Exam Free Online Quiz 2020, CAT Full Online Mock Test Series 4th in English. CAT Online Test for All Subjects, CAT Free Mock Test Series in English. CAT Free Mock Test Series 4. CAT English Language Online Test in English Series 4th. Here we are providing CAT Full Mock Test Paper in English. CAT Mock Test Series 4th 2020. Now Test your self for CAT Exam by using below quiz…
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A batsman in his 17^{th} innings makes a score of 85 and their by increasing his average by 3. What is his average after the 17^{th}innings?
16x + 85 = 17(x + 3)
x = 34 + 3 = 37
16x + 85 = 17(x + 3)
x = 34 + 3 = 37
The price of a VCR is marked at Rs. 12,000. If successive discounts of 15%, 10% and 5% be allowed, then at what price does a customer buy it?
Actual price = 95% of 90% of 85% of Rs. 12000
= 95/100 * 90/100 * 85/100 * 12000 = Rs. 8721.
Actual price = 95% of 90% of 85% of Rs. 12000
= 95/100 * 90/100 * 85/100 * 12000 = Rs. 8721.
The base of a right triangle is 8 and hypotenuse is 10. Its area is?
108 * 107 * 96 = ? (to the nearest hundred)
108 * 107 * 96 ≡ 1109400
108 * 107 * 96 ≡ 1109400
Two pipes can separately fill a tank in 20 and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is full, a leak develops in the tank through which onethird of water supplied by both the pipes goes out. What is the total time taken to fill the tank?
1/20 + 1/30 = 1/12
1 + 1/3 = 4/3
1 — 12
4/3 — ?
4/3 * 12 = 16 hrs
1/20 + 1/30 = 1/12
1 + 1/3 = 4/3
1 — 12
4/3 — ?
4/3 * 12 = 16 hrs
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr. x – 5 = 45 ==> x = 50 km/hr.
Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr. x – 5 = 45 ==> x = 50 km/hr.
5 * 5 ÷ 5 + 5 ÷ 5 = ?
(5 * 5)/5 + 1/5 * 5 = 5 + 1 = 6
(5 * 5)/5 + 1/5 * 5 = 5 + 1 = 6
I. a^{2} + 11a + 30 = 0,
II. b^{2} + 6b + 5 = 0 to solve both the equations to find the values of a and b?
I. (a + 6)(a + 5) = 0
=> a = 6, 5
II. (b + 5)(b + 1) = 0
=> b = 5, 1 => a ≤ b
I. (a + 6)(a + 5) = 0
=> a = 6, 5
II. (b + 5)(b + 1) = 0
=> b = 5, 1 => a ≤ b
9000 + 16 2/3 % of ? = 10500
9000 + 16 2/3 % of ? = 10500 => 9000 + 50/3 % of ? = 10500
50/(3 * 100) of ? = 1500 => ? = 1500 * 6
? = 9000
9000 + 16 2/3 % of ? = 10500 => 9000 + 50/3 % of ? = 10500
50/(3 * 100) of ? = 1500 => ? = 1500 * 6
? = 9000
9 3/4 + 7 2/17 – 9 1/15 = ?
Given sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)
= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)
= 7 + (765 + 120 – 68)/1020 = 7 817/1020
Given sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)
= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)
= 7 + (765 + 120 – 68)/1020 = 7 817/1020
The number of new words that can be formed by rearranging the letters of the word ‘ALIVE’ is .
Number of words which can be formed = 5! – 1 = 120 – 1 = 119.
Number of words which can be formed = 5! – 1 = 120 – 1 = 119.
(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓
(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓
=> (6 ÷ 2 ÷ 9) * ? = 100 / 3
=> 3 / 9 * ? = 100 / 3
=> ? = 100
(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓
=> (6 ÷ 2 ÷ 9) * ? = 100 / 3
=> 3 / 9 * ? = 100 / 3
=> ? = 100
A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.
Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.
If 35% of a number is 12 less than 50% of that number, then the number is?
Let the number be x. Then,
50% of x – 35% of x = 12
50/100 x – 35/100 x = 12
x = (12 * 100)/15 = 80.
Let the number be x. Then,
50% of x – 35% of x = 12
50/100 x – 35/100 x = 12
x = (12 * 100)/15 = 80.
The sum of the digits of a twodigit number is 12. The difference of the digits is 6. Find the number?
Let the twodigit number be 10a + b
a + b = 12 — (1)
If a>b, a – b = 6
If b>a, b – a = 6
If a – b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 – a = 3
Number would be 93.
if b – a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 – b = 3.
Number would be 39.
There fore, Number would be 39 or 93.
Let the twodigit number be 10a + b
a + b = 12 — (1)
If a>b, a – b = 6
If b>a, b – a = 6
If a – b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 – a = 3
Number would be 93.
if b – a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 – b = 3.
Number would be 39.
There fore, Number would be 39 or 93.
The ratio of the prices of three articles X, Y and Z is 8 : 5 : 3. If the prices of X , Y and Z are increased by 25%, 20% and 33 1/3% respectively, then what would be the ratio of the new prices of X, Y and Z?
Let the prices of X, Y and Z be 8k, 5k and 3k respectively.
After increase
Price of X = 8k * 125/100 = 10k
Price of Y = 5k * 120/100 = 6k
Price of Z = 3k * (133 1/3)/100 = 4k
Required ratio = 10k : 6k : 4k = 5 : 3 : 2.
Let the prices of X, Y and Z be 8k, 5k and 3k respectively.
After increase
Price of X = 8k * 125/100 = 10k
Price of Y = 5k * 120/100 = 6k
Price of Z = 3k * (133 1/3)/100 = 4k
Required ratio = 10k : 6k : 4k = 5 : 3 : 2.
Find the least multiple of 13 which when divided by 6, 8 and 12 leaves 5, 7 and 11 as remainders respectively?
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in?
(A + B + C)’s 1 day work = 1/6;
(A + B)’s 1 day work = 1/8
(B + C)’s 1 day work = 1/12
(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)
= (1/3 – 5/24) = 1/8
So, A and C together will do the work in 8 days.
(A + B + C)’s 1 day work = 1/6;
(A + B)’s 1 day work = 1/8
(B + C)’s 1 day work = 1/12
(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)
= (1/3 – 5/24) = 1/8
So, A and C together will do the work in 8 days.
If the height of a cone is increased by 100% then its volume is increased by?
The ratio of the radius of two circles is 1: 3, and then the ratio of their areas is?
r_{1}: r_{2} = 1: 3
Πr_{1}^{2}: Πr_{2}^{2}
r_{1}^{2}: r_{2}^{2 }= 1: 9
r_{1}: r_{2} = 1: 3
Πr_{1}^{2}: Πr_{2}^{2}
r_{1}^{2}: r_{2}^{2 }= 1: 9
There is a 30% increase in the price of an article in the first year, a 20% decrease in the second year and a 10% increase in the next year. If the final price of the article is Rs. 2288, then what was the price of the article initially?
Let the price of the article, four years age be Rs. 100 in the 1st year, price of the article = 100 + 30 = Rs. 130. In the 2nd year, price = 130 – 20% of 130 = 130 – 26 = Rs. 104.
In the 3rd year, price = 104 + 10% of 104 = 104 + 10.4 = Rs. 114.40.
But present price of the article is Rs. 2288
for 114.4 —> 100 ; 2288 —> ?
Required price = (2288 * 100)/114.4 = 20 * 100 = Rs. 2000.
Let the price of the article, four years age be Rs. 100 in the 1st year, price of the article = 100 + 30 = Rs. 130. In the 2nd year, price = 130 – 20% of 130 = 130 – 26 = Rs. 104.
In the 3rd year, price = 104 + 10% of 104 = 104 + 10.4 = Rs. 114.40.
But present price of the article is Rs. 2288
for 114.4 —> 100 ; 2288 —> ?
Required price = (2288 * 100)/114.4 = 20 * 100 = Rs. 2000.
Difference between two numbers is 5, six times of the smaller lacks by 6 from the four times of the greater. Find the numbers?
x – y = 5
4x – 6y = 6
x = 12 y = 7
x – y = 5
4x – 6y = 6
x = 12 y = 7
The sum of a number and its reciprocal is oneeighth of 34. What is the product of the number and its square root?
Let the number be x. Then,
x + 1/x = 34/8
8x^{2} – 34x + 8 = 0
4x^{2} – 17x + 4 = 0
(4x – 1)(x – 4) = 0
x = 4
required number = 4 * √4 = 4 * 2 = 8.
Let the number be x. Then,
x + 1/x = 34/8
8x^{2} – 34x + 8 = 0
4x^{2} – 17x + 4 = 0
(4x – 1)(x – 4) = 0
x = 4
required number = 4 * √4 = 4 * 2 = 8.
The sum of the digits of a twodigit number is 15 and the difference between the digits is 3. What is the twodigit number?
Let the ten’s digit be x and unit’s digit by y
Then, x + y = 15 and x – y = 3 or y – x = 3
Solving x + y = 15 and x – y = 3, we get: x = 9, y = 6
Solving x + y = 15 and y – x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69.
Hence, the number cannot be determined.
Let the ten’s digit be x and unit’s digit by y
Then, x + y = 15 and x – y = 3 or y – x = 3
Solving x + y = 15 and x – y = 3, we get: x = 9, y = 6
Solving x + y = 15 and y – x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69.
Hence, the number cannot be determined.
The length of a rectangle is increased by 25% and its breadth is decreased by 20%. What is the effect on its area?
100 * 100 = 10000
125 * 80 = 10000
No change
100 * 100 = 10000
125 * 80 = 10000
No change
A and B can finish a work in 16 days while A alone can do the same work in 24 days. In how many days B alone will complete the work?
B = 1/16 – 1/24 = 1/48 => 48 days
B = 1/16 – 1/24 = 1/48 => 48 days
The ratio of the earnings of P and Q is 9 : 10. If the earnings of P increases by onefourth and the earnings of Q decreases by onefourth, then find the new ratio of their earnings?
Let the earnings of P and Q be Rs. 9x and Rs. 10x respectively.
New ration = [9x + 1/4(9x)]/[10x – 1/4(10x)]
=> [9x(1 + 1/4)]/[10x(1 – 1/4)] = 9/10 * (5/4)/(3/4) => 3/2.
Let the earnings of P and Q be Rs. 9x and Rs. 10x respectively.
New ration = [9x + 1/4(9x)]/[10x – 1/4(10x)]
=> [9x(1 + 1/4)]/[10x(1 – 1/4)] = 9/10 * (5/4)/(3/4) => 3/2.
The food in a camp lasts for 30 men for 40 days. If ten more men join, how many days will the food last?
one man can consume the same food in 30*40 = 1200 days.
10 more men join, the total number of men = 40
The number of days the food will last = 1200/40 = 30 days.
one man can consume the same food in 30*40 = 1200 days.
10 more men join, the total number of men = 40
The number of days the food will last = 1200/40 = 30 days.
The roots of the equation 3x^{2} – 12x + 10 = 0 are?
The discriminant of the quadratic equation is (12)^{2} – 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.
The discriminant of the quadratic equation is (12)^{2} – 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.
If one third of 3/4 of a number is 21. Then find the number?
x * 1/3 * 3/4 =21 => x = 84
x * 1/3 * 3/4 =21 => x = 84
A, B and C can do a work in 6, 8 and 12 days respectively doing the work together and get a payment of Rs.1800. What is B’s share?
WC = 1/6:1/8:1/12 => 4:3:2
3/9 * 1800 = 600
WC = 1/6:1/8:1/12 => 4:3:2
3/9 * 1800 = 600
Two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction?
Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/5 = 8 m/sec.
Relative speed = 12 + 8 = 20 m/sec.
Required time = (120 + 120)/20 = 12 sec.
Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/5 = 8 m/sec.
Relative speed = 12 + 8 = 20 m/sec.
Required time = (120 + 120)/20 = 12 sec.
The owner of a furniture shop charges his customer 24% more than the cost price. If a customer paid Rs. 8339 for a computer table, then what was the cost price of the computer table?
CP = SP * (100/(100 + profit%))
= 8339(100/124) = Rs. 6725.
CP = SP * (100/(100 + profit%))
= 8339(100/124) = Rs. 6725.
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m.
Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m.
Six points are marked on a straight line and five points are marked on another line which is parallel to the first line. How many straight lines, including the first two, can be formed with these points?
We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear.
Hence, the required number of straight lines
= ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1
= 55 – 15 – 10 + 2 = 32
We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear.
Hence, the required number of straight lines
= ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1
= 55 – 15 – 10 + 2 = 32
The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 10, then their difference is:
Let the numbers be x and (100 – x).
Then, x(100 – x) = 5 * 495
x^{2} – 100x + 2475 = 0
(x – 55)(x – 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 – 45 = 10.
Let the numbers be x and (100 – x).
Then, x(100 – x) = 5 * 495
x^{2} – 100x + 2475 = 0
(x – 55)(x – 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 – 45 = 10.
(743.30)^{2} = ?
(743.30)^{2} = 552500
(743.30)^{2} = 552500
9 men and 12 boys finish a job in 12 days, 12 men and 12 boys finish it in 10 days. 10 men and 10 boys shall finish it in how many days?
9M + 12B —– 12 days
12M + 12B —— 10 days
10M + 10B ——?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B
18B + 12B = 30B — 12 days
20B + 10B = 30B —–? => 12 days
9M + 12B —– 12 days
12M + 12B —— 10 days
10M + 10B ——?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B
18B + 12B = 30B — 12 days
20B + 10B = 30B —–? => 12 days
The difference between a number and its threefifth is 50. What is the number?
Let the number be x. Then,
x – 3/5 x = 50 => 2/5 x = 50
x = (50 * 5)/2 = 125.
Let the number be x. Then,
x – 3/5 x = 50 => 2/5 x = 50
x = (50 * 5)/2 = 125.
If the cost price of 50 articles is equal to the selling price of 40 articles, then the gain or loss percent is?
Given that, cost price of 50 article is equal to selling price of 40 articles.
Let cost price of one article = Rs. 1
Selling price of 40 articles = Rs. 50
But Cost price of 40 articles = Rs. 40
Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25%
Given that, cost price of 50 article is equal to selling price of 40 articles.
Let cost price of one article = Rs. 1
Selling price of 40 articles = Rs. 50
But Cost price of 40 articles = Rs. 40
Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25%
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train.
Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9 = 150 meter
Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9 = 150 meter
A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?
90%
104%
——–
14% — 140
100% — ? => Rs.1000
90%
104%
——–
14% — 140
100% — ? => Rs.1000
Sum of two numbers is 80. Greater number exceeds by 5 from four times of the smaller. Find the numbers?
x + y = 80
x – 4y = 5
x = 65 y = 15
x + y = 80
x – 4y = 5
x = 65 y = 15
The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?
Let the rate of interest be R% p.a.
4400{[1 + R/100]^{2} – 1} = 11193.60
[1 + R/100]^{2} = (44000 + 11193.60)/44000
[1 + R/100]^{2} = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]^{2} = 784/625 = (28/25)^{2}
1 + R/100 = 28/25
R/100 = 3/25
herefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560
Let the rate of interest be R% p.a.
4400{[1 + R/100]^{2} – 1} = 11193.60
[1 + R/100]^{2} = (44000 + 11193.60)/44000
[1 + R/100]^{2} = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]^{2} = 784/625 = (28/25)^{2}
1 + R/100 = 28/25
R/100 = 3/25
herefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560
If the L.C.M of two numbers is 750 and their product is 18750, find the H.C.F of the numbers.
H.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.
H.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.
Albert buys 4 horses and 9 cows for Rs. 13,400. If he sells the horses at 10% profit and the cows at 20% profit, then he earns a total profit of Rs. 1880. The cost of a horse is:
Let C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.
Then, 4x + 9y = 13400 — (i)
And, 10% of 4x + 20% of 9y = 1880
2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)
Solving (i) and (ii), we get : x = 2000 and y = 600.
Cost price of each horse = Rs. 2000.
Let C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.
Then, 4x + 9y = 13400 — (i)
And, 10% of 4x + 20% of 9y = 1880
2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)
Solving (i) and (ii), we get : x = 2000 and y = 600.
Cost price of each horse = Rs. 2000.
Robert is traveling on his cycle and has calculated to reach point A at 2 p.m. if he travels at 10 km/hr; he will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach A at 1 p.m.?
Let the distance traveled be x km.
Then, x/10 – x/15 = 2
3x – 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = 60/5 = 12 kmph.
Let the distance traveled be x km.
Then, x/10 – x/15 = 2
3x – 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = 60/5 = 12 kmph.
A certain sum is invested at simple interest at 18% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 840. Find the sum?
Let the sum be Rs. x.
(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840
=> 12x/100 = 840 => x = 7000.
Let the sum be Rs. x.
(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840
=> 12x/100 = 840 => x = 7000.
A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 – 6 = 6 liters
Remaining water = 8 – 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 – 8) = 8 liters.
Remaining water = (4 2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 – 6 = 6 liters
Remaining water = 8 – 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 – 8) = 8 liters.
Remaining water = (4 2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.
How much interest can a person get on Rs. 8200 at 17.5% p.a. simple interest for a period of two years and six months?
I = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50
I = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50
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The average of five results is 46 and that of the first four is 45. The fifth result is?
5 * 46 – 4 * 45 = 50
5 * 46 – 4 * 45 = 50
Which of the following is the greatest?
5 > 4 and 1/2 > 1/4
5^{1/2} > 4^{1/4}
4^{1/4} cannot be the greatest. Hence 3^{1/4} also cannot be the greatest.
3^{7/10} = (3^{7})^{1/10} = (2187)^{1/10}
5^{1/2} = 5^{5/10} = (5^{5})^{1/10} = (3125)^{1/10}
6^{1/5} = 6^{2/10} = (6^{2})^{1/10} = (36)^{1/10}
As (3125)^{1/10} > (2187)^{1/10} > (36)^{1/10} , 5^{1/2} is the greatest.
5 > 4 and 1/2 > 1/4
5^{1/2} > 4^{1/4}
4^{1/4} cannot be the greatest. Hence 3^{1/4} also cannot be the greatest.
3^{7/10} = (3^{7})^{1/10} = (2187)^{1/10}
5^{1/2} = 5^{5/10} = (5^{5})^{1/10} = (3125)^{1/10}
6^{1/5} = 6^{2/10} = (6^{2})^{1/10} = (36)^{1/10}
As (3125)^{1/10} > (2187)^{1/10} > (36)^{1/10} , 5^{1/2} is the greatest.
A tank is filled in eight hours by three pipes A, B and C. Pipe A is twice as fast as pipe B, and B is twice as fast as C. How much time will pipe B alone take to fill the tank?
1/A + 1/B + 1/C = 1/8 (Given)
Also given that A = 2B and B = 2C
=> 1/2B + 1/B + 2/B = 1/8
=> (1 + 2 + 4)/2B = 1/8
=> 2B/7 = 8
=> B = 28 hours.
1/A + 1/B + 1/C = 1/8 (Given)
Also given that A = 2B and B = 2C
=> 1/2B + 1/B + 2/B = 1/8
=> (1 + 2 + 4)/2B = 1/8
=> 2B/7 = 8
=> B = 28 hours.
Amar, Bhavan and Chetan divide an amount of Rs.5600 among themselves in the ratio 3:6:5. If an amount of Rs.400 is deducted from each of their shares, what will be the new ratio of their shares of the amount?
Let the shares of Amar, Bhavan and Chetan be 3x, 6x and 5x respectively.
3x + 6x + 5x = 5600
14x = 5600 => x = 400
Required ratio = 3x – 400 : 6x – 400 : 5x – 400
= 3x – x : 6x – x : 5x – x
= 2x : 5x : 4x => 2:5:4
Let the shares of Amar, Bhavan and Chetan be 3x, 6x and 5x respectively.
3x + 6x + 5x = 5600
14x = 5600 => x = 400
Required ratio = 3x – 400 : 6x – 400 : 5x – 400
= 3x – x : 6x – x : 5x – x
= 2x : 5x : 4x => 2:5:4
A can finish a piece of work in 5 days. B can do it in 10 days. They work together for two days and then A goes away. In how many days will B finish the work?
2/5 + (2 + x)/10 = 1 => x = 4 days
2/5 + (2 + x)/10 = 1 => x = 4 days
The length of rectangle is thrice its breadth and its perimeter is 96 m, find the area of the rectangle?
2(3x + x) = 96
l = 36 b = 12
lb = 36 * 12 = 432
2(3x + x) = 96
l = 36 b = 12
lb = 36 * 12 = 432
Find the one which does not belong to that group ?
In each of the groups B4, D16, E25 and F36 the number is the square of the position value of the letter. This pattern is not followed in I91.
In each of the groups B4, D16, E25 and F36 the number is the square of the position value of the letter. This pattern is not followed in I91.
The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is:
Excluded number = (27 * 5) – (25 * 4) = 135 – 100 = 35.
Excluded number = (27 * 5) – (25 * 4) = 135 – 100 = 35.
When 2 is added to half of onethird of onefifth of a number, the result is onefifteenth of the number. Find the number?
Let the number be
2 + 1/2[1/3(a/5)] = a/15
=> 2 = a/30 => a = 60
Let the number be
2 + 1/2[1/3(a/5)] = a/15
=> 2 = a/30 => a = 60
In an office, totally there are 6400 employees and 65% of the total employees are males. 25% of the males in the office are atleast 50 years old. Find the number of males aged below 50 years?
Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 – 25)%
= 4160 * 75/100 = 3120.
Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 – 25)%
= 4160 * 75/100 = 3120.
A train crosses a platform of 150 m in 15 sec, same train crosses another platform of length 250 m in 20 sec. then find the length of the train?
Length of the train be ‘X’
X + 150/15 = X + 250/20
4X + 600 = 3X + 750
X = 150m
Length of the train be ‘X’
X + 150/15 = X + 250/20
4X + 600 = 3X + 750
X = 150m
A shopkeeper sells two articles at Rs.1000 each, making a profit of 20% on the first article and a loss of 20% on the second article. Find the net profit or loss that he makes?
SP of first article = 1000
Profit = 20%
CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3
SP of second article = 1000
Loss = 20%
CP = (SP)*[100/(100L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CPSP = (6250/3) 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100 =
0.04 * 100 = 4%
SP of first article = 1000
Profit = 20%
CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3
SP of second article = 1000
Loss = 20%
CP = (SP)*[100/(100L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CPSP = (6250/3) 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100 =
0.04 * 100 = 4%
Pipe A can fill a tank in 6 hours. Due to a leak at the bottom, it takes 9 hours for the pipe A to fill the tank. In what time can the leak alone empty the full tank?
Let the leak can empty the full tank in x hours 1/6 – 1/x = 1/9
=> 1/x = 1/6 – 1/9 = (3 – 2)/18 = 1/18
=> x = 18.
Let the leak can empty the full tank in x hours 1/6 – 1/x = 1/9
=> 1/x = 1/6 – 1/9 = (3 – 2)/18 = 1/18
=> x = 18.
A and B can do a work in 12 days, B and C in 15 days, C and A in 20 days. If A, B and C work together, they will complete the work in?
(A + B)’s 1 day work = 1/12;
(B + C)’s 1 day work = 1/15;
(A + C)’s 1 day work = 1/20
Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5
(A + B + C)’s 1 day work = 1/10
So, A, B and C together can complete the work in 10 days.
(A + B)’s 1 day work = 1/12;
(B + C)’s 1 day work = 1/15;
(A + C)’s 1 day work = 1/20
Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5
(A + B + C)’s 1 day work = 1/10
So, A, B and C together can complete the work in 10 days.
Find the greatest 4digit number exactly divisible by 3, 4 and 5?
Greatest 4digit number is 9999. LCM of 3, 4 and 5 is 60. Dividing 9999 with 60, we get remainder 39. Thus, the required number is 9999 – 39 = 9960.
Greatest 4digit number is 9999. LCM of 3, 4 and 5 is 60. Dividing 9999 with 60, we get remainder 39. Thus, the required number is 9999 – 39 = 9960.
A boy rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately?
Total distance traveled = 10 + 12 = 22 km /hr.
Total time taken = 10/12 + 12/10 = 61/30 hrs.
Average speed = 22 * 30/61 = 10.8 km/hr.
Total distance traveled = 10 + 12 = 22 km /hr.
Total time taken = 10/12 + 12/10 = 61/30 hrs.
Average speed = 22 * 30/61 = 10.8 km/hr.
The average age of three boys is 15 years and their ages are in proportion 3:5:7. What is the age in years of the youngest boy?
3x + 5x + 7x = 45
x =3
3x = 9
3x + 5x + 7x = 45
x =3
3x = 9
Difference between two numbers is 5, six times of the smaller lacks by 6 from the four times of the greater. Find the numbers?
x – y = 5
4x – 6y = 6
x = 12 y = 7
x – y = 5
4x – 6y = 6
x = 12 y = 7
(6 – 2 * 3 + 6) / (2 + 6 * 3 – 4) = ?
(6 – 2 * 3 + 6) / (2 + 6 * 3 – 4) = (6 – 6 + 6) / (2 + 18 – 4)
= 6/16 = 3/8
(6 – 2 * 3 + 6) / (2 + 6 * 3 – 4) = (6 – 6 + 6) / (2 + 18 – 4)
= 6/16 = 3/8
The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average of the team?
Let the average of the whole team be x years.
11x – (26 + 29) = 9(x – 1)
= 11x – 9x = 46
= 2x = 46 => x = 23
So, average age of the team is 23 years.
Let the average of the whole team be x years.
11x – (26 + 29) = 9(x – 1)
= 11x – 9x = 46
= 2x = 46 => x = 23
So, average age of the team is 23 years.
31.2 * 14.5 * 9.6 = ?
? = 4343.04
? = 4343.04
A man whose speed is 4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph, find his average speed for the total journey?
M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9 = 4
M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9 = 4
A train 150 m long running at 72 kmph crosses a platform in 25 sec. What is the length of the platform?
D = 72 * 5/18 = 25 = 500 – 150 = 350
D = 72 * 5/18 = 25 = 500 – 150 = 350
Find the greatest number that, while dividing 47, 215 and 365, gives the same remainder in each case?
Calculate the differences, taking two numbers at a time as follows:
(21547) = 168
(365215) = 150
(36547) = 318
HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each cases is 5.
Calculate the differences, taking two numbers at a time as follows:
(21547) = 168
(365215) = 150
(36547) = 318
HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each cases is 5.
David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B?
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves onefourth of his income, find the ratio of their monthly savings?
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.
What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high?
d^{2} = 12^{2} + 4^{2} + 3^{2} = 13
d^{2} = 12^{2} + 4^{2} + 3^{2} = 13
The spherical ball of lead 3 cm in diameter is melted and recast into 3 spherical balls. The diameters of two of these are 1 ½ cm and 2 cm respectively. The diameter of third ball is?
4/3 π * 3 * 3 * 3 = 4/3 π[(3/2)^{3} + 2^{3} + r^{3}]
r = 1.25
d = 2.5
4/3 π * 3 * 3 * 3 = 4/3 π[(3/2)^{3} + 2^{3} + r^{3}]
r = 1.25
d = 2.5
In a division sum, the remainder is 0. As student mistook the divisor by 12 instead of 21 and obtained 35 as quotient. What is the correct quotient?
Number = (12 * 35) = 420
Correct quotient = 420/21 = 20
Number = (12 * 35) = 420
Correct quotient = 420/21 = 20
The denominator of a fraction is 1 less than twice the numerator. If the numerator and denominator are both increased by 1, the fraction becomes 3/5. Find the fraction?
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9.
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9.
Sum of two numbers is 80. Greater number exceeds by 5 from four times of the smaller. Find the numbers?
x + y = 80
x – 4y = 5
x = 65 y = 15
x + y = 80
x – 4y = 5
x = 65 y = 15
Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full is?
(A + B)’s 1 hour work = (1/12 + 1/15) = 3/20
(A + C)’s 1 hour work = (1/12 + 1/20) = 2/15
Part filled in 2 hrs = (3/20 + 2/15) = 17/60
Part filled in 6 hrs = 3 * 17/60 = 17/20
Remaining part = 1 – 17/20 = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
Total time taken to fill the tank = (6 + 1) = 7 hrs.
(A + B)’s 1 hour work = (1/12 + 1/15) = 3/20
(A + C)’s 1 hour work = (1/12 + 1/20) = 2/15
Part filled in 2 hrs = (3/20 + 2/15) = 17/60
Part filled in 6 hrs = 3 * 17/60 = 17/20
Remaining part = 1 – 17/20 = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
Total time taken to fill the tank = (6 + 1) = 7 hrs.
The ratio between the present ages of P and Q is 6:7. If Q is 4 years old than P, what will be the ratio of the ages of P and Q after 4 years?
Let P’s age and Q’s age be 6x and 7x years respectively.
Then, 7x – 6x = 4 => x = 4
Required ratio = (6x + 4) : (7x + 4)
28 : 32 = 7:8
Let P’s age and Q’s age be 6x and 7x years respectively.
Then, 7x – 6x = 4 => x = 4
Required ratio = (6x + 4) : (7x + 4)
28 : 32 = 7:8
A, B, C enter into a partnership investing Rs. 35,000, Rs. 45,000 and Rs. 55,000 respectively. The respective shares of A, B, C in annual profit of Rs. 40,500 are:
A:B:C = 35000 : 45000 : 55000 = 7:9:11
A’s share = 40500 * 7/27 = Rs. 10500
B’s share = 40500 * 9/27 = Rs. 13500
C’s share = 40500 * 11/27 = Rs. 16500
A:B:C = 35000 : 45000 : 55000 = 7:9:11
A’s share = 40500 * 7/27 = Rs. 10500
B’s share = 40500 * 9/27 = Rs. 13500
C’s share = 40500 * 11/27 = Rs. 16500
If (4^{61} + 4^{62} + 4^{63} + 4^{64}) is divisible by ?, then ? =
4^{61} + 4^{62} + 4^{63} + 4^{64}
=> 4^{61}(1 + 4 + 16 + 64)
4^{61}(85) = 17 (5 * 4^{61}) So it is divisible by 17.
4^{61} + 4^{62} + 4^{63} + 4^{64}
=> 4^{61}(1 + 4 + 16 + 64)
4^{61}(85) = 17 (5 * 4^{61}) So it is divisible by 17.
A money lender finds that due to a fall in the annual rate of interest from 8% to 7 3/4 % his yearly income diminishes by Rs. 61.50, his capital is?
Let the capital be Rs. x. Then,
(x * 8 * 1)/100 – (x * 31/4 * 1/100) = 61.50
32x – 31x = 6150 * 4
x = 24,600.
Let the capital be Rs. x. Then,
(x * 8 * 1)/100 – (x * 31/4 * 1/100) = 61.50
32x – 31x = 6150 * 4
x = 24,600.
Find the nearest to 25268 which is exactly divisible by 467?
The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?
Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 —– (1)
(10Q + P) – (10P + Q) = 54
9(Q – P) = 54
(Q – P) = 6 —– (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28
Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 —– (1)
(10Q + P) – (10P + Q) = 54
9(Q – P) = 54
(Q – P) = 6 —– (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28
50 men can complete a work in 65 days.Five days after started the work, 20 men left the group. In how many days can the remaining work be completed?
After 5 days, the following situation prevails.
50 men can complete the work in 60 days.
30 men can complete the work in ? days.
M_{1} D_{1} = M_{2} D_{2}
=> 50 * 60 = 30 * D_{2}
=> D_{2} = (50 * 60)/30 = 100 days.
After 5 days, the following situation prevails.
50 men can complete the work in 60 days.
30 men can complete the work in ? days.
M_{1} D_{1} = M_{2} D_{2}
=> 50 * 60 = 30 * D_{2}
=> D_{2} = (50 * 60)/30 = 100 days.
A rectangular field 30 m long and 20 m broad. How much deep it should be dug so that from the earth taken out, a platform can be formed which is 8 m long, 5.5 m broad and 1.5 m high where as the earth taken out is increase by 10/5?
30 * 20 * x = (8 * 5.5 * 1.5)/2
30 * 20 * x = (8 * 5.5 * 1.5)/2
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8) = ?
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8)
= (0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / 2^{3}(0.8 * 0.8 * 0.8 + 0.9 * 0.9 * 0.9) = 1/8 = 0.125
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8)
= (0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / 2^{3}(0.8 * 0.8 * 0.8 + 0.9 * 0.9 * 0.9) = 1/8 = 0.125
A man can row a boat at 20 kmph in still water. If the speed of the stream is 6 kmph, what is the time taken to row a distance of 60 km downstream?
Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 60/26 = 30/13 hours.
Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 60/26 = 30/13 hours.
The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19, 29 and 34 respectively is?
LCM = 1400
1400 – 6 = 1394
LCM = 1400
1400 – 6 = 1394
Rajan borrowed Rs.4000 at 5% p.a compound interest. After 2 years, he repaid Rs.2210 and after 2 more year, the balance with interest. What was the total amount that he paid as interest?
4000
200 — I
200
10 — II
—————
4410
2210
——–
2000
110 — III
110
5.50 — IV
———–
2425.50
2210
———–
4635.50
4000
———
635.50
4000
200 — I
200
10 — II
—————
4410
2210
——–
2000
110 — III
110
5.50 — IV
———–
2425.50
2210
———–
4635.50
4000
———
635.50
The average runs scored by a batsman in 20 matches is 40. In the next 10 matches the batsman scored an average of 13 runs. Find his average in all the 30 matches?
Total score of the batsman in 20 matches = 800.
Total score of the batsman in the next 10 matches = 130.
Total score of the batsman in the 30 matches = 930.
Average score of the batsman = 930/30 = 31.
Total score of the batsman in 20 matches = 800.
Total score of the batsman in the next 10 matches = 130.
Total score of the batsman in the 30 matches = 930.
Average score of the batsman = 930/30 = 31.
Kiran travels from A to B by car and returns from B to A by cycle in 7 hours. If he travels both ways by car he saves 3 hours. What is the time taken to cover both ways by cycle?
Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.
x + y = 7 — (1) ; 2x = 4 — (2)
solving both the equations, we get y = 5
So, time taken to cover both ways by cycle = 2y hours = 10 hours.
Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.
x + y = 7 — (1) ; 2x = 4 — (2)
solving both the equations, we get y = 5
So, time taken to cover both ways by cycle = 2y hours = 10 hours.
The area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm is?
(5 * 3.5)/2 = 8.75
(5 * 3.5)/2 = 8.75
David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B?
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
How many meters of carpet 50cm, wide will be required to cover the floor of a room 30m * 20m?
50/100 * x = 30 * 20 => x = 1200
50/100 * x = 30 * 20 => x = 1200
3 men, 4 women and 6 children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days?
Let 1 woman’s 1 day work = x.
Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.
So, (3x/2 + 4x + + 6x/4) = 1/7
28x/4 = 1/7 => x = 1/49
1 woman alone can complete the work in 49 days.
So, to complete the work in 7 days, number of women required = 49/7 = 7.
Let 1 woman’s 1 day work = x.
Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.
So, (3x/2 + 4x + + 6x/4) = 1/7
28x/4 = 1/7 => x = 1/49
1 woman alone can complete the work in 49 days.
So, to complete the work in 7 days, number of women required = 49/7 = 7.
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A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 4^{5} – 1.
= 2^{10} – 1 = 1024 – 1 = 1023
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 4^{5} – 1.
= 2^{10} – 1 = 1024 – 1 = 1023
The denominator of a fraction is 1 less than twice the numerator. If the numerator and denominator are both increased by 1, the fraction becomes 3/5. Find the fraction?
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9.
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9.
The number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places?
The word MEADOWS has 7 letters of which 3 are vowels.
VVV
As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.
Hence the total ways are 24 * 6 = 144.
The word MEADOWS has 7 letters of which 3 are vowels.
VVV
As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.
Hence the total ways are 24 * 6 = 144.
64 boys and 40 girls form a group for social work. During their membership drive, same number of boys and girls joined the group. How many members does the group have now, if the ratio of boys to girls is 4 : 3?
Let us say x boys and x girls joined the group.
(64 + x) / (40 + x) = 4/3
192 + 3x = 160 + 4x => x = 32
Number of members in the group = 64 + x + 40 + x
= 104 + 2x = 168.
Let us say x boys and x girls joined the group.
(64 + x) / (40 + x) = 4/3
192 + 3x = 160 + 4x => x = 32
Number of members in the group = 64 + x + 40 + x
= 104 + 2x = 168.
The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 867 sq m, then what is the breadth of the rectangular plot?
Let the breadth of the plot be b m.
Length of the plot = 3 b m
(3b)(b) = 867
3b^{2} = 867
b^{2} = 289 = 17^{2} (b > 0)
b = 17 m.
Let the breadth of the plot be b m.
Length of the plot = 3 b m
(3b)(b) = 867
3b^{2} = 867
b^{2} = 289 = 17^{2} (b > 0)
b = 17 m.
Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?
2 * 22/7 * 14 = 88
88 * 1 1/2 = Rs.132
2 * 22/7 * 14 = 88
88 * 1 1/2 = Rs.132
A and B can do a piece of work in 6 2/3 days and 5 days respectively. They work together for 2 days and then A leaves. In how many days after that B will complete the work alone.
3/20 * 2 + (2 + x)/5 = 1
x = 1 ½ days
3/20 * 2 + (2 + x)/5 = 1
x = 1 ½ days
Find the one which does not belong to that group ?
Except in 862, in all other numbers sum of first two digits is same as the last digit.
Except in 862, in all other numbers sum of first two digits is same as the last digit.
Simplify the following:
(169/121)^{3/2} * 27/2 * (13/22)^{1}
(169/121)^{3/2} * 27/2 * (13/22)^{1}
(13^{2}/11^{2})^{3/2} * 27/2 * 22/13 = 13^{3}/11^{3} * 27/2 * 22/13
= 13^{4} 11^{4} 3^{3}
(169/121)^{3/2} * 27/2 * (13/22)^{1}
(13^{2}/11^{2})^{3/2} * 27/2 * 22/13 = 13^{3}/11^{3} * 27/2 * 22/13
= 13^{4} 11^{4} 3^{3}
How much time will a train of length 200 m moving at a speed of 72 kmph take to cross another train of length 300 m, moving at 36 kmph in the same direction?
The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.
Relative speed = 72 36 = 36 kmph = 36 * 5/18 = 10 mps.
Time required = d/s = 500/10 = 50 sec.
The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.
Relative speed = 72 36 = 36 kmph = 36 * 5/18 = 10 mps.
Time required = d/s = 500/10 = 50 sec.
(56^{2} – 24^{2}) * 1/32 + ?% of 1200 = 146
(56^{2} – 24^{2}) * 1/32 + ?% of 1200 = 146
=> (3136 – 576) * 1/32 + ?/32 of 1200 = 146
=> 2560/32 + ? * 12 = 146 => ? * 12 = 146 – 80
=> ? = 66/12 => 5 1/2
(56^{2} – 24^{2}) * 1/32 + ?% of 1200 = 146
=> (3136 – 576) * 1/32 + ?/32 of 1200 = 146
=> 2560/32 + ? * 12 = 146 => ? * 12 = 146 – 80
=> ? = 66/12 => 5 1/2
The profit earned by selling an article for Rs. 832 is equal to the loss incurred when the same article is sold for Rs. 448. What should be the sale price for making 50% profit?
Let C.P. = Rs. x.
Then, 832 – x = x – 448
2x = 1280 => x = 640
Required S.P. = 150% of Rs. 640 = 150/100 * 640 = Rs. 960.
Let C.P. = Rs. x.
Then, 832 – x = x – 448
2x = 1280 => x = 640
Required S.P. = 150% of Rs. 640 = 150/100 * 640 = Rs. 960.
A train running at a speed of 36 kmph crosses an electric pole in 12 seconds. In how much time will it cross a 350 m long platform?
Let the length of the train be x m.
When a train crosses an electric pole, the distance covered is its own length.
So, x = 12 * 36 * 5 /18 m = 120 m.
Time taken to cross the platform = (120 +350)/ 36 * 5/18 = 47 min.
Let the length of the train be x m.
When a train crosses an electric pole, the distance covered is its own length.
So, x = 12 * 36 * 5 /18 m = 120 m.
Time taken to cross the platform = (120 +350)/ 36 * 5/18 = 47 min.
(3^{3} * 9^{5/2}) / (27^{2/3} * 3^{4}) = ?
(3^{3} * 9^{5/2}) / (27^{2/3} * 3^{4}) = [3^{3} * (3^{2})^{5/2}] / [(3^{3})^{2/3} * 3^{4}]
= (3^{3} * 3^{5}) / (3^{2} * 3^{4}) = (3^{3+5})/(3^{24}) = 3^{2}/3^{2} = 3^{2} * 3^{2} = 81.
(3^{3} * 9^{5/2}) / (27^{2/3} * 3^{4}) = [3^{3} * (3^{2})^{5/2}] / [(3^{3})^{2/3} * 3^{4}]
= (3^{3} * 3^{5}) / (3^{2} * 3^{4}) = (3^{3+5})/(3^{24}) = 3^{2}/3^{2} = 3^{2} * 3^{2} = 81.
The average runs scored by a batsman in 20 matches is 40. In the next 10 matches the batsman scored an average of 13 runs. Find his average in all the 30 matches?
Total score of the batsman in 20 matches = 800.
Total score of the batsman in the next 10 matches = 130.
Total score of the batsman in the 30 matches = 930.
Average score of the batsman = 930/30 = 31.
Total score of the batsman in 20 matches = 800.
Total score of the batsman in the next 10 matches = 130.
Total score of the batsman in the 30 matches = 930.
Average score of the batsman = 930/30 = 31.
If 3 workers collect 48 kg of cotton in 4 days, how many kg of cotton will 9 workers collect in 2 days?
(3 * 4)/48 = (9 * 2)/ x
x = 72 kg
(3 * 4)/48 = (9 * 2)/ x
x = 72 kg
In a 1000 m race, A beats B by 50 m and B beats C by 100 m. In the same race, by how many meters does A beat C?
By the time A covers 1000 m, B covers (1000 – 50) = 950 m.
By the time B covers 1000 m, C covers (1000 – 100) = 900 m.
So, the ratio of speeds of A and C =
1000/950 * 1000/900 = 1000/855 So, by the time A covers 1000 m, C covers 855 m.
So in 1000 m race A beats C by 1000 – 855 = 145 m.
By the time A covers 1000 m, B covers (1000 – 50) = 950 m.
By the time B covers 1000 m, C covers (1000 – 100) = 900 m.
So, the ratio of speeds of A and C =
1000/950 * 1000/900 = 1000/855 So, by the time A covers 1000 m, C covers 855 m.
So in 1000 m race A beats C by 1000 – 855 = 145 m.
A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior?
The total number of ways of forming the group of ten representatives is ²²C₁₀.
The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way
The required number of ways = ²²C₁₀ – 1
The total number of ways of forming the group of ten representatives is ²²C₁₀.
The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way
The required number of ways = ²²C₁₀ – 1
I. a^{3} – 988 = 343,
II. b^{2} – 72 = 49 to solve both the equations to find the values of a and b?
a^{3} = 1331 => a = 11
b^{2} = 121 => b = ± 11
a ≥ b
a^{3} = 1331 => a = 11
b^{2} = 121 => b = ± 11
a ≥ b
The slant height of a cone is 12 cm and radius of the base is 4cm, find the curved surface of the cone.
π * 12 * 4 = 48
π * 12 * 4 = 48
A wire in the form of a circle of radius 3.5 m is bent in the form of a rectangule, whose length and breadth are in the ratio of 6 : 5. What is the area of the rectangle?
The circumference of the circle is equal to the permeter of the rectangle.
Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5
=> x = 1
Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm^{2}
The circumference of the circle is equal to the permeter of the rectangle.
Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5
=> x = 1
Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm^{2}
Find the one which does not belong to that group ?
Skin, Eye, Nose and Ear are sensory organs, while Leg is a limb.
Skin, Eye, Nose and Ear are sensory organs, while Leg is a limb.
A plot ABCD is as shown in figure, where AF = 30 m, CE = 40 m, ED = 50 m, AE = 120 m. Find the area of the plot ABCD?
Area of plot ABCD = Area of ADE + Area of AFB + Area of BCEF = 1/2 * 50 * 120 + 1/2 * 40 * 30 + 40 * 90 = 3000 + 600 + 3600 = 7200 sq.m
Area of plot ABCD = Area of ADE + Area of AFB + Area of BCEF = 1/2 * 50 * 120 + 1/2 * 40 * 30 + 40 * 90 = 3000 + 600 + 3600 = 7200 sq.m
The average of first 10 prime numbers is?
Sum of 10 prime no. = 129
Average = 129/10 = 12.9
Sum of 10 prime no. = 129
Average = 129/10 = 12.9
? % of 400 + 40% of 160 = 17% of 400
? % of 400 + 40% of 160 = 17% of 400
?/100 of 400 + 40/100 of 160 = 17/100 of 400
?(4) + 64 = 68 => ? = 1
? % of 400 + 40% of 160 = 17% of 400
?/100 of 400 + 40/100 of 160 = 17/100 of 400
?(4) + 64 = 68 => ? = 1
A laborer is engaged for 30 days on the condition that he receives Rs.25 for each day he works and is fined Rs.7.50 for each day is absent. He gets Rs.425 in all. For how many days was he absent?
30 * 25 = 750
425
———–
325
25 + 7.50 = 32.5
325/32.5 = 10
30 * 25 = 750
425
———–
325
25 + 7.50 = 32.5
325/32.5 = 10
A and B starts a business with Rs.8000 each, and after 4 months, B withdraws half of his capital . How should they share the profits at the end of the 18 months?
A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 4 months and then withdraws Rs.4000. So, the investment of B for remaining 14 months is Rs.4000 only.
A : B
8000*18 : (8000*4) + (4000*14)
14400 : 88000
A:B = 18:11
A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 4 months and then withdraws Rs.4000. So, the investment of B for remaining 14 months is Rs.4000 only.
A : B
8000*18 : (8000*4) + (4000*14)
14400 : 88000
A:B = 18:11
If the sum of onehalf of a number exceeds onethird of that number by 7 1/3, the number is:
Let the number be x. Then,
(1/2 x + 1/5 x) – 1/3 x = 22/3
11/30 x = 22/3
x = 20
Let the number be x. Then,
(1/2 x + 1/5 x) – 1/3 x = 22/3
11/30 x = 22/3
x = 20
What is the difference between the largest number and the least number written with the digits 7, 3, 1, 4?
1347
7431
————
6084
1347
7431
————
6084
The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?
Let the sides of the rectangle be l and b respectively.
From the given data,
(√l^{2} + b^{2}) = (1 + 108 1/3 %)lb
=> l^{2} + b^{2} = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l^{2} + b^{2})/lb = 25/12
12(l^{2} + b^{2}) = 25lb
Adding 24lb on both sides
12l^{2} + 12b^{2} + 24lb = 49lb
12(l^{2} + b^{2} + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)^{2} = 49lb
=> 12(14)^{2} = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.
Let the sides of the rectangle be l and b respectively.
From the given data,
(√l^{2} + b^{2}) = (1 + 108 1/3 %)lb
=> l^{2} + b^{2} = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l^{2} + b^{2})/lb = 25/12
12(l^{2} + b^{2}) = 25lb
Adding 24lb on both sides
12l^{2} + 12b^{2} + 24lb = 49lb
12(l^{2} + b^{2} + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)^{2} = 49lb
=> 12(14)^{2} = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.
The inverse ratio of 3: 2: 1 is?
1/3: 1/2: 1/1 = 2:3:6
1/3: 1/2: 1/1 = 2:3:6
The wheels revolve round a common horizontal axis. They make 15, 20 and 48 revolutions in a minute respectively. Starting with a certain point on the circumference down wards. After what interval of time will they come together in the same position?
Time for one revolution = 60/15 = 4
60/ 20 = 3
60/48 = 5/4
LCM of 4, 3, 5/4
LCM of Numerators/HCF of Denominators =
60/1 = 60
Time for one revolution = 60/15 = 4
60/ 20 = 3
60/48 = 5/4
LCM of 4, 3, 5/4
LCM of Numerators/HCF of Denominators =
60/1 = 60
If x % of 80 = 20% of y, then x = ? and y = ?
x % of 80 = 20 % of y => x / 100 (80) = 20 / 100 (y)
=> 4x = y
i.e x / y = ¼
x % of 80 = 20 % of y => x / 100 (80) = 20 / 100 (y)
=> 4x = y
i.e x / y = ¼
The distance between Delhi and Mathura is 110 kms. A starts from Delhi with a speed of 20 kmph at 7 a.m. for Mathura and B starts from Mathura with a speed of 25 kmph at 8 p.m. from Delhi. When will they meet?
D = 110 – 20 = 90
RS = 20 + 25 = 45
T = 90/45 = 2 hours
8 a.m. + 2 = 10 a.m.
D = 110 – 20 = 90
RS = 20 + 25 = 45
T = 90/45 = 2 hours
8 a.m. + 2 = 10 a.m.
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 sec. The speed of the train is?
Speed of the train relative to man = 125/10 = 25/2 m/sec.
= 25/2 * 18/5 = 45 km/hr
Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr.
x – 5 = 45 => x = 50 km/hr.
Speed of the train relative to man = 125/10 = 25/2 m/sec.
= 25/2 * 18/5 = 45 km/hr
Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr.
x – 5 = 45 => x = 50 km/hr.
The average weight of a group of persons increased from 48 kg to 51 kg, when two persons weighing 78 kg and 93 kg join the group. Find the initial number of members in the group?
Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n – 48n = 69 => n = 23
Therefore there were 23 members in the group initially.
Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n – 48n = 69 => n = 23
Therefore there were 23 members in the group initially.
Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely?
D = 250 m + 250 m = 500 m
RS = 80 + 70 = 150 * 5/18 = 125/3
T = 500 * 3/125 = 12 sec
D = 250 m + 250 m = 500 m
RS = 80 + 70 = 150 * 5/18 = 125/3
T = 500 * 3/125 = 12 sec
If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is?
Let the actual distance traveled be x km. Then,
x/10 = (x + 20)/14
4x – 200 =>x = 50 km.
Let the actual distance traveled be x km. Then,
x/10 = (x + 20)/14
4x – 200 =>x = 50 km.
The weights of three boys are in the ratio 4 : 5 : 6. If the sum of the weights of the heaviest and the lightest boy is 45 kg more than the weight of the third boy, what is the weight of the lightest boy?
Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy
= 4k = 4(9) = 36 kg.
Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy
= 4k = 4(9) = 36 kg.
The speed of a train is 90 kmph. What is the distance covered by it in 10 minutes?
90 * 10/60 = 15 kmph
90 * 10/60 = 15 kmph
If p, q and r are positive integers and satisfy x = (p + q r)/r = (p – q + r)/q = (q + r – p)/p, then the value of x is?
When two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is nonzero.
Hence, x = (p + q r)/r = (p – q + r)/q = (q + r – p)/p
=> x = (p + q – r + p – q + r + q + r – p) / (r + q + p)
=> x = (r + q + p) / (r + q + p) = 1
p + q + r is nonzero.
When two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is nonzero.
Hence, x = (p + q r)/r = (p – q + r)/q = (q + r – p)/p
=> x = (p + q – r + p – q + r + q + r – p) / (r + q + p)
=> x = (r + q + p) / (r + q + p) = 1
p + q + r is nonzero.
Find the one which does not belong to that group ?
Crocodile, Turtle, Allegator and Frog are amphibians, while Chameleon is a terrestrial animal.
Crocodile, Turtle, Allegator and Frog are amphibians, while Chameleon is a terrestrial animal.
A train 150 m long running at 72 kmph crosses a platform in 25 sec. What is the length of the platform?
D = 72 * 5/18 = 25 = 500 – 150 = 350
D = 72 * 5/18 = 25 = 500 – 150 = 350
64 is what percent of 80?
Let x percent of 80 be 64.
80 * x/100 = 64 => x = (64 * 100)/80 => x = 80.
80% of 80 is 64.
Let x percent of 80 be 64.
80 * x/100 = 64 => x = (64 * 100)/80 => x = 80.
80% of 80 is 64.
A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m?
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000
The sum of five consecutive odd numbers of set p is 435. What is the sum of five consecutive numbers of another set q. Whose largest number is 45 more than the largest number of set p?
Let the five consecutive odd numbers of set p be 2n – 3, 2n – 1, 2n + 1, 2n + 3, 2n + 5.
Sum of these five numbers
= 2n – 3 + 2n – 1 + 2n + 1 + 2n + 3 + 2n + 5
= 10n + 5 = 435 => n = 43
Largest number of set p = 2(43) + 5 = 91
The largest number of set q = 91 + 45 = 136
=> The five numbers of set q are 132, 133, 134, 135, 136.
Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670.
Let the five consecutive odd numbers of set p be 2n – 3, 2n –