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If the sides of a cube are in the ratio 4:3. What is the ratio of their diagonals?
a_{1}:a_{2} = 4:3
d_{1}:d_{2} = 4:3
a_{1}:a_{2} = 4:3
d_{1}:d_{2} = 4:3
Find the one which does not belong to that group ?
Y^{+4}C^{2}A^{+4}E^{2}C, K^{+4}O^{2}M^{+4}Q^{2}O, P^{+4}T^{2}R^{+3}U^{1}T, G^{+4}K^{2}I^{+4}M^{2}K and D^{+4}H^{2}F^{+4}J^{2}H.
Except PTRUT, all other groups follow similar pattern.
Y^{+4}C^{2}A^{+4}E^{2}C, K^{+4}O^{2}M^{+4}Q^{2}O, P^{+4}T^{2}R^{+3}U^{1}T, G^{+4}K^{2}I^{+4}M^{2}K and D^{+4}H^{2}F^{+4}J^{2}H.
Except PTRUT, all other groups follow similar pattern.
A man sells two articles for Rs.3600 each and he gains 30% on the first and loses 30% on the next. Find his total gain or loss?
(30*30)/100 = 9%loss
(30*30)/100 = 9%loss
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.
I. a^{2} – 7a + 12 = 0,
II. b^{2} – 3b + 2 = 0 to solve both the equations to find the values of a and b?
I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > b
I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > b
If an article is sold at 19% profit instead of 12% profit, then the profit would be Rs. 105 more. What is the cost price?
Let the cost price of an article be Rs. x.
(19% of x) – (12% of x) = 105
19x/100 – 12x/100 = 105 => 7x = 105 * 100
=> x = 1500
Cost price = Rs. 1500
Let the cost price of an article be Rs. x.
(19% of x) – (12% of x) = 105
19x/100 – 12x/100 = 105 => 7x = 105 * 100
=> x = 1500
Cost price = Rs. 1500
(64 + 9 + 9) / (2 * 20 + 1) = ?
(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2
(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2
All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k – 5k = 3k.
Quantity of water in container C = 8k – 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k – 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k – 5k = 3k.
Quantity of water in container C = 8k – 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k – 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.
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
The least number of five digits that have 144 their HCF is?
144) 10000 (69
9936
——–
64
10000 + 144 – 64 = 10080
144) 10000 (69
9936
——–
64
10000 + 144 – 64 = 10080
x varies inversely as square of y. Given that y = 2 for x = 1. The value of x for y = 6 will be equal to:
Given x = k/y^{2}, where k is a constant.
Now, y = 2 and x = 1 gives k = 4.
x = 4/y^{2} => x = 4/6^{2}, when
y = 6 => x = 4/36 = 1/9.
Given x = k/y^{2}, where k is a constant.
Now, y = 2 and x = 1 gives k = 4.
x = 4/y^{2} => x = 4/6^{2}, when
y = 6 => x = 4/36 = 1/9.
If the area of circle is 616 sq cm then its circumference?
22/7 r^{2} = 616 => r = 14
2 * 22/7 * 14 = 88
22/7 r^{2} = 616 => r = 14
2 * 22/7 * 14 = 88
In a bag there are coins of 50 paisa, 25 paisa and one rupee in the proportion 5:6:2. If there are in all Rs.42, the number of 25 paisa coins is?
5x 6x 2x
50 25 100
250x + 150x + 200x = 4200
600x = 4200
x = 7 => 6x = 42
5x 6x 2x
50 25 100
250x + 150x + 200x = 4200
600x = 4200
x = 7 => 6x = 42
A can finish a work in 18 days B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?
B’s 10 day’s work = 1/15 * 10 = 2/3
Remaining work = (1 – 2/3) = 1/3
Now, 1/18 work is done by A in 1 day.
1/3 work is done by A in (18 * 1/3) = 6 days.
B’s 10 day’s work = 1/15 * 10 = 2/3
Remaining work = (1 – 2/3) = 1/3
Now, 1/18 work is done by A in 1 day.
1/3 work is done by A in (18 * 1/3) = 6 days.
A, B and C completed a piece of work, A worked for 6 days, B for 9 days and C for 4 days. Their daily wages were in the ratio of 3:4:5. Find the daily wages of C, if their total earning was Rs.1480?
3x 4x 5x
6 9 4
18x + 36x + 20x = 1480
74x = 1480 => x = 20
5x = 100 Rs.
3x 4x 5x
6 9 4
18x + 36x + 20x = 1480
74x = 1480 => x = 20
5x = 100 Rs.
7394 – 1236 – 4652 = ?
If x;Y = 5:2, then (8x + 9y):(8x + 2y) is :
Let x = 5k and y = 2k.
Then, (8x + 9y)/(8x + 2y)
= [(8 * 5k) + (9 * 2k)] / [(8 * 5k) + (2 * 2k)] = 58k/44k = 29/22
(8x + 9y):(8x + 2y) = 29:22
Let x = 5k and y = 2k.
Then, (8x + 9y)/(8x + 2y)
= [(8 * 5k) + (9 * 2k)] / [(8 * 5k) + (2 * 2k)] = 58k/44k = 29/22
(8x + 9y):(8x + 2y) = 29:22
A student scored an average of 80 marks in 3 subjects: Physics, Chemistry and Mathematics. If the average marks in Physics and Mathematics is 90 and that in Physics and Chemistry is 70, what are the marks in Physics?
Given M + P + C = 80 * 3 = 240 — (1)
M + P = 90 * 2 = 180 — (2)
P + C = 70 * 2 = 140 — (3)
Where M, P and C are marks obtained by the student in Mathematics, Physics and Chemistry.
P = (2) + (3) – (1) = 180 + 140 – 240 = 80
Given M + P + C = 80 * 3 = 240 — (1)
M + P = 90 * 2 = 180 — (2)
P + C = 70 * 2 = 140 — (3)
Where M, P and C are marks obtained by the student in Mathematics, Physics and Chemistry.
P = (2) + (3) – (1) = 180 + 140 – 240 = 80
Find the nearest to 25268 which is exactly divisible by 467?
What is the difference between the compound interest on Rs.12000 at 20% p.a. for one year when compounded yearly and half yearly?
When compounded annually, interest
= 12000[1 + 20/100]^{1} – 12000 = Rs.2400
When compounded semiannually, interest
= 12000[1 + 10/100]^{2} – 12000 = Rs.2520
Required difference = 2520 – 2400 = Rs.120
When compounded annually, interest
= 12000[1 + 20/100]^{1} – 12000 = Rs.2400
When compounded semiannually, interest
= 12000[1 + 10/100]^{2} – 12000 = Rs.2520
Required difference = 2520 – 2400 = Rs.120
How many three letter words are formed using the letters of the word TIME?
The number of letters in the given word is four.
The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.
The number of letters in the given word is four.
The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.
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.
If a: b = 3:4, b:c = 7:9, c:d = 5:7, find a:d?
a/d = (3/4)*(7/9)*(5/7) => 5/12
a/d = (3/4)*(7/9)*(5/7) => 5/12
Solve for m if 36 (6^{m}) = (216)^{3m+4}
36 (6^{m}) = (216)^{3m+4} => 6^{2} 6^{m} = (6^{3})^{3m+4} => 6^{2+m} = 6^{9m+12} Equating powers of 6 on both sides.
m + 2 = 9m + 12 => 10 = 8m => m = 5/4
36 (6^{m}) = (216)^{3m+4} => 6^{2} 6^{m} = (6^{3})^{3m+4} => 6^{2+m} = 6^{9m+12} Equating powers of 6 on both sides.
m + 2 = 9m + 12 => 10 = 8m => m = 5/4
X men can do a work in 120 days. If there were 20 men less, the work would have taken 60 days more. What is the value of X?
We have M_{1} D_{1} = M_{2} D_{2}
120X = (X – 20)180
=> 2X = (X – 20) 3 => 2X = 3X – 60
=> X = 60
We have M_{1} D_{1} = M_{2} D_{2}
120X = (X – 20)180
=> 2X = (X – 20) 3 => 2X = 3X – 60
=> X = 60
Which of the following groups of fractions is in descending order?
The fractions considered are 8/15 9/13 6/11
To compare them we make the denominators the same. So the fractions are
(8 * 13 * 11)/2145, (9 * 15 * 11)/2145 and (6 * 15 * 13)/2145
1144/2145, 1485/2145 and 1170/2145
so in descending order the fractions will be
1485/2145, 1170/2145 and 1144/2145 i.e., 9/13 , 6/11 , 8/15
The fractions considered are 8/15 9/13 6/11
To compare them we make the denominators the same. So the fractions are
(8 * 13 * 11)/2145, (9 * 15 * 11)/2145 and (6 * 15 * 13)/2145
1144/2145, 1485/2145 and 1170/2145
so in descending order the fractions will be
1485/2145, 1170/2145 and 1144/2145 i.e., 9/13 , 6/11 , 8/15
The least number which when diminished by 7 is divisible by 21, 28, 36 and 45 is?
LCM = 1260
1260 + 7 = 1267
LCM = 1260
1260 + 7 = 1267
A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?
A runs 1000 m while B runs 900 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= (1000 * 800)/900 = 8000/9 = 888.88 m.
B can give C = 1000 – 888.88 = 111.12 m.
A runs 1000 m while B runs 900 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= (1000 * 800)/900 = 8000/9 = 888.88 m.
B can give C = 1000 – 888.88 = 111.12 m.
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.
If the price has fallen by 10% what percent of its consumption be: increased so that the expenditure may be the same as before?
100 – 10 = 90
90——10
100——? => 11 1/9%
100 – 10 = 90
90——10
100——? => 11 1/9%
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.
Two trains of equal length, running with the speeds of 60 and 40 kmph, take 50 seconds to cross each other while they are running in the same direction. What time will they take to cross each other if they are running in opposite directions?
RS = 60 40 = 20 * 5/18 = 100/18
T = 50
D = 50 * 100/18 = 2500/9
RS = 60 + 40 = 100 * 5/18
T = 2500/9 * 18/500 = 10 sec
RS = 60 40 = 20 * 5/18 = 100/18
T = 50
D = 50 * 100/18 = 2500/9
RS = 60 + 40 = 100 * 5/18
T = 2500/9 * 18/500 = 10 sec
The average salary of workers in an industry is Rs.200 the average salary of technicians being Rs.400 and that of nontechnicians being Rs.125. What is the total number of workers?
A can do a half of certain work in 70 days and B one third of the same in 35 days. They together will do the whole work in.
A = 140 days
B = 105 days
1/140 + 1/105 = 7/420 = 1/60
=>60 days
A = 140 days
B = 105 days
1/140 + 1/105 = 7/420 = 1/60
=>60 days
It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is?
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120/x + 480/y = 8 or 1/x + 4/y = 1/15 — (i)
And, 200/x + 400/y = 25/3 or 1/x + 2/y = 1/24 — (ii)
Solving (i) and (ii), we get x = 60 and y = 80
Ratio of speeds = 60:80 = 3:4
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120/x + 480/y = 8 or 1/x + 4/y = 1/15 — (i)
And, 200/x + 400/y = 25/3 or 1/x + 2/y = 1/24 — (ii)
Solving (i) and (ii), we get x = 60 and y = 80
Ratio of speeds = 60:80 = 3:4
A man purchases 8 pens for Rs.9 and sells 9 pens for Rs.8, how much profit or loss does he make?
81 — 17
100 —– ? è 20.98%loss
81 — 17
100 —– ? è 20.98%loss
A and B can do a piece of work in 3 days, B and C in 4 days, C and A in 6 days. How long will C take to do it?
2c = ¼ + 1/6 – 1/3 = 1/12
c = 1/24 => 24 days
2c = ¼ + 1/6 – 1/3 = 1/12
c = 1/24 => 24 days
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.
Find the expenditure on digging a well 14m deep and of 3m diameter at Rs.15 per cubic meter?
22/7 * 14 * 3/2 * 3/2 = 99 m^{2}
99 * 15 = 1485
22/7 * 14 * 3/2 * 3/2 = 99 m^{2}
99 * 15 = 1485
[50 – {20 + 60% of (26/13 + 24/12 + 21) – 2} + 8] = 625/?
[50 – {20 + 60% of (26/13 + 24/12 + 21) – 2} + 8] = 625/x => [50 – {20 + 3/5 of (25) – 2} + 8] = 625/x
=> [50 – 33 + 8] = 625/x => x = 25
[50 – {20 + 60% of (26/13 + 24/12 + 21) – 2} + 8] = 625/x => [50 – {20 + 3/5 of (25) – 2} + 8] = 625/x
=> [50 – 33 + 8] = 625/x => x = 25
The average marks in mathematics scored by the pupils of a school at the public examination were 39. If four of these pupils who actually scored 5, 12, 15 and 19 marks at the examination had not been sent up, the average marks for the school would have been 44. Find the number of pupils sent up for examination from the school?
39x = 5 + 12 + 15 + 19 + (x – 4)44
x = 25
39x = 5 + 12 + 15 + 19 + (x – 4)44
x = 25
Surface area of two spheres are in the ratio 1:4 what is the ratio of their volumes?
A tank 3 m long, 2 m wide and 1.5 m deep is dug in a field 20 m long and 14 m wide. If the earth dug out is evenly spread out over the field, the level of the field will raise by nearly?
The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers?
Let the numbers x, x + 2, x + 4, x + 6 and x + 8
Then, [x + (x + 2) + (x + 4) + (x + 6) + (x + 8)]5 = 61
or 5x + 20 = 305 => x = 57
So, required difference = (57 + 8) – 57 = 8.
Let the numbers x, x + 2, x + 4, x + 6 and x + 8
Then, [x + (x + 2) + (x + 4) + (x + 6) + (x + 8)]5 = 61
or 5x + 20 = 305 => x = 57
So, required difference = (57 + 8) – 57 = 8.
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%
The least number which when divided by 8, 12 and 16, leave in each remainder 3, is?
LCM = 48 + 3 = 51
LCM = 48 + 3 = 51
(0.9 * 0.9 – 0.8 * 0.8)/1.7 = ?
As numerator is of the form a^{2} – b^{2} = (a + b) (a – b), so the given expression simplifying becomes
(0.9^{2} – 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1
As numerator is of the form a^{2} – b^{2} = (a + b) (a – b), so the given expression simplifying becomes
(0.9^{2} – 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1
If a number is chosen at random from the set {1, 2, 3, …., 100}, then the probability that the chosen number is a perfect cube is .
We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
4/100 = 1/25.
We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
4/100 = 1/25.
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.
Ravi can do a piece of work in 30 days while Prakash can do it in 40 days. In how many days will they finish it together?
1/30 + 1/40 = 7/120
120/7 = 17 1/7 days
1/30 + 1/40 = 7/120
120/7 = 17 1/7 days
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The simple interest accrued on an amount Rs.10,000 at the end of two years is same as the compound interest on Rs.8,000 at the end of two years. The rate of interest is same in both the cases. What is the rate of interest?
Given that Rs.10,000 is invested in S.I for two years and Rs.8,000 in C.I for two years
=> C.I – S.I
=> 8,000 { [1 + r/100]^{2 } – 1} = (10,000)2r /100
=> 8{ 1 + 2r/100 + r^{2} / (100)^{2} – 1} = r/5
=> 16r/100 + 8r^{2}/(100)^{2} = 20r/100
=> 4r/10 = 8r^{2}/(100)^{2}
=> 8[r/100]^{2} – 4r/100 = 0
=> r/100 {8r/100 4} = 0
=> r = 0% of 50%
Since r!= 0%, r =50%
Given that Rs.10,000 is invested in S.I for two years and Rs.8,000 in C.I for two years
=> C.I – S.I
=> 8,000 { [1 + r/100]^{2 } – 1} = (10,000)2r /100
=> 8{ 1 + 2r/100 + r^{2} / (100)^{2} – 1} = r/5
=> 16r/100 + 8r^{2}/(100)^{2} = 20r/100
=> 4r/10 = 8r^{2}/(100)^{2}
=> 8[r/100]^{2} – 4r/100 = 0
=> r/100 {8r/100 4} = 0
=> r = 0% of 50%
Since r!= 0%, r =50%
The smallest number when increased by ” 1 ” is exactly divisible by 12, 18, 24, 32 and 40 is:
LCM = 1440
1440 – 1 = 1439
LCM = 1440
1440 – 1 = 1439
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
In a can, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can?
Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T – 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 – 8 = 4/9(T – 8)
10T = 792 – 352 => T = 44.
Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T – 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 – 8 = 4/9(T – 8)
10T = 792 – 352 => T = 44.
421 * 0.9 + 130 * 101 + 10000 = ?
421 * 0.9 + 130 * 101 + 10000
= 378.9 + 1313 + 10000 = 23508.9 = 23500
421 * 0.9 + 130 * 101 + 10000
= 378.9 + 1313 + 10000 = 23508.9 = 23500
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.
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
Twofifth of onethird of threeseventh of a number is 15. What is 40% of that number?
Let the number be x. Then,
2/5 of 1/3 of 3/7 of
x = 15 => x = (15 * 5/3 * 3 * 5/2) = 525/2
40% of 525/2 = (40/100 * 525/2) = 105
Let the number be x. Then,
2/5 of 1/3 of 3/7 of
x = 15 => x = (15 * 5/3 * 3 * 5/2) = 525/2
40% of 525/2 = (40/100 * 525/2) = 105
A, B and C invests Rs.2000, Rs.3000 and Rs.4000 in a business. After one year A removed his money; B and C continued the business for one more year. If the net profit after 2 years be Rs.3200, then A’s share in the profit is?
2*12 : 3*12 : 4*24
1: 3: 4
1/8 * 3200 = 400
2*12 : 3*12 : 4*24
1: 3: 4
1/8 * 3200 = 400
The difference between simple and compound interest on Rs. 1200 for one year at 10% per annum reckoned halfyearly is?
S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)^{2} – 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3.
S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)^{2} – 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3.
H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 2^{4} * 3^{5} * 5^{2} * 7^{2}. The third number is:
3240 = 2^{3} * 3^{4} * 5; 3600 = 2^{4} * 3^{2} * 5^{2}
H.C.F = 36 = 2^{2} * 3^{2}
Since H.C.F is the product of lowest powers of common factors, so the third number must have (2^{2} * 3^{2}) as its factor.
Since L.C.M is the product of highest powers of common prime factors, so the third number must have 3^{5} and 7^{2} as its factors.
Third number = 2^{2} * 3^{5} * 7^{2}
3240 = 2^{3} * 3^{4} * 5; 3600 = 2^{4} * 3^{2} * 5^{2}
H.C.F = 36 = 2^{2} * 3^{2}
Since H.C.F is the product of lowest powers of common factors, so the third number must have (2^{2} * 3^{2}) as its factor.
Since L.C.M is the product of highest powers of common prime factors, so the third number must have 3^{5} and 7^{2} as its factors.
Third number = 2^{2} * 3^{5} * 7^{2}
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 9x and 10x respectively.
New ratio = [9x + 1/4 (9x)]/[10x – 1/4 (10x)]
=> 9*(1 + 1/4)/10*(1 – 1/4)
=> 9/10 * (5/4)/(3/4) = 3/2
Let the earnings of P and Q be 9x and 10x respectively.
New ratio = [9x + 1/4 (9x)]/[10x – 1/4 (10x)]
=> 9*(1 + 1/4)/10*(1 – 1/4)
=> 9/10 * (5/4)/(3/4) = 3/2
I. x^{2} + 9x + 20 = 0,
II. y^{2} + 5y + 6 = 0 to solve both the equations to find the values of x and y?
I. x^{2} + 4x + 5x + 20 = 0
=>(x + 4)(x + 5) = 0 => x = 4, 5
II. y^{2} + 3y + 2y + 6 = 0
=>(y + 3)(y + 2) = 0 => y = 3, 2
= x < y.
I. x^{2} + 4x + 5x + 20 = 0
=>(x + 4)(x + 5) = 0 => x = 4, 5
II. y^{2} + 3y + 2y + 6 = 0
=>(y + 3)(y + 2) = 0 => y = 3, 2
= x < y.
LCM of 455, 117, 338 is:
The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 42 kmph, find the speed of the stream?
The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(21) = 3/1 = 3:1
Speed of the stream = 42/3 = 14 kmph.
The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(21) = 3/1 = 3:1
Speed of the stream = 42/3 = 14 kmph.
A person takes 20 minutes more to cover a certain distance by decreasing his speed by 20%. What is the time taken to cover the distance at his original speed?
Let the distance and original speed be d km and k kmph respectively.
d/0.8k – d/k = 20/60 => 5d/4k – d/k = 1/3
=> (5d – 4d)/4k = 1/3 => d = 4/3 k
Time taken to cover the distance at original speed
= d/k = 4/3 hours = 1 hour 20 minutes.
Let the distance and original speed be d km and k kmph respectively.
d/0.8k – d/k = 20/60 => 5d/4k – d/k = 1/3
=> (5d – 4d)/4k = 1/3 => d = 4/3 k
Time taken to cover the distance at original speed
= d/k = 4/3 hours = 1 hour 20 minutes.
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
A gardener wants to plant trees in his garden in rows in such away that the number of trees in each row to be the same. If there are 24 rows the number of trees in each row is 42 if there are 12 more rows find the number of trees in each row?
Required number of trees
= 24/36 * 42 = 28.
Required number of trees
= 24/36 * 42 = 28.
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
A cistern has three pipes, A, B and C. The pipes A and B can fill it in 4 and 5 hours respectively and C can empty it in 2 hours. If the pipes are opened in order at 1, 2 and 3 A.M. When will the cistern be empty?
1 to 2 = 1/4
2 to 3 = 1/4 + 1/5 = 9/20
After 3 AM = 1/4 + 1/5 – 1/2 = 1/20
1/4 + 9/20 = 14/20
1 h — 1/20
? —– 14/20
14 hours ==> 5 PM
1 to 2 = 1/4
2 to 3 = 1/4 + 1/5 = 9/20
After 3 AM = 1/4 + 1/5 – 1/2 = 1/20
1/4 + 9/20 = 14/20
1 h — 1/20
? —– 14/20
14 hours ==> 5 PM
Find the sum The difference between the compound and S.I. on a certain sum of money for 2 years at 10% per annum is Rs.15of money?
P = 15(100/10)^{2 }=> P = 1500
P = 15(100/10)^{2 }=> P = 1500
A person got Rs.48 more when he invested a certain sum at compound interest instead of simple interest for two years at 8% p.a. Find the sum?
P = (d * 100^{2}) / R^{2}
=> (48 * 100 * 100) / 8 * 8 = Rs.7500
P = (d * 100^{2}) / R^{2}
=> (48 * 100 * 100) / 8 * 8 = Rs.7500
How much 60% of 50 is greater than 40% of 30?
(60/100) * 50 – (40/100) * 30
30 – 12 = 18
(60/100) * 50 – (40/100) * 30
30 – 12 = 18
A train 100 m long crosses a platform 125 m long in 15 sec; find the speed of the train?
D = 100 + 125 = 225
T = 15
S = 225/15 * 18/5 = 54 kmph
D = 100 + 125 = 225
T = 15
S = 225/15 * 18/5 = 54 kmph
The length of the bridge, which a train 130 m long and traveling at 45 km/hr can cross in 30 sec is?
Speed = 45 * 5/18 = 25/2 m/sec.
Time = 30 sec
Let the length of bridge be x meters.
Then, (130 + x)/30 = 25/2
x = 245 m.
Speed = 45 * 5/18 = 25/2 m/sec.
Time = 30 sec
Let the length of bridge be x meters.
Then, (130 + x)/30 = 25/2
x = 245 m.
Rs.525 among A, B and C such that B may get 2/3 of A and C together get. Find the share of C?
From a pack of cards two cards are drawn one after the other, with replacement. The probability that the first is a red card and the second is a king is .
Let E_{1} be the event of drawing a red card.
Let E_{2} be the event of drawing a king .
P(E_{1} ∩ E_{2}) = P(E_{1}) . P(E_{2})
(As E_{1} and E_{2} are independent)
= 1/2 * 1/13 = 1/26
Let E_{1} be the event of drawing a red card.
Let E_{2} be the event of drawing a king .
P(E_{1} ∩ E_{2}) = P(E_{1}) . P(E_{2})
(As E_{1} and E_{2} are independent)
= 1/2 * 1/13 = 1/26
Eight men, ten women and six boys together can complete a piece of work in eight days. In how many days can 20 women complete the same work if 20 men can complete it in 12 days?
Let the number of units which can be completed by each man, each women and each boy be m/day, w/day and b/day respectively.
Work = 8(8m + 10w +6b)units = 12(20m)
10w +6b = 22m
b is unknown.
We cannot find the relation between m and w.
We cannot answer the question.
Let the number of units which can be completed by each man, each women and each boy be m/day, w/day and b/day respectively.
Work = 8(8m + 10w +6b)units = 12(20m)
10w +6b = 22m
b is unknown.
We cannot find the relation between m and w.
We cannot answer the question.
the present average age of a couple and their daughter is 35 years. Fifteen years from now, the age of the mother will be equal to the sum of present ages of the father and the daughter. Find the present age of mother?
(f + m + d)/3 = 35
=> f + m + d = 105 — (1)
m + 15 = f + d
Substituting f + d as m + 15 in (1), we get
2m + 15 = 105
2m = 90 => m = 45 years.
(f + m + d)/3 = 35
=> f + m + d = 105 — (1)
m + 15 = f + d
Substituting f + d as m + 15 in (1), we get
2m + 15 = 105
2m = 90 => m = 45 years.
A man sells two articles for Rs.3600 each and he gains 30% on the first and loses 30% on the next. Find his total gain or loss?
(30*30)/100 = 9%loss
(30*30)/100 = 9%loss
96 is divided into two parts in such a way that seventh part of first and ninth part of second are equal. Find the smallest part?
x/7 = y/9 => x:y = 7:9
7/16 * 96 = 42
x/7 = y/9 => x:y = 7:9
7/16 * 96 = 42
Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm^{2}
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm^{2}
The average age of 15 students of a class is 15 years. Out of these, the average age of 5 students is 14 years and that of the other 9 students is 16 years. Tee age of the 15th student is:
Age of the 15th student=
[15 * 15 – (14 * 5 + 16 * 9)] = (225 – 214) = 11 years.
Age of the 15th student=
[15 * 15 – (14 * 5 + 16 * 9)] = (225 – 214) = 11 years.
A, B and C can do a piece of work in 7 days, 14 days and 28 days respectively. How long will they taken, if all the three work together?
1/7 + 1/14 + 1/28 = 7/28 = 1/4 => 4 days
1/7 + 1/14 + 1/28 = 7/28 = 1/4 => 4 days
Five bells first begin to toll together and then at intervals of 5, 10, 15, 20 and 25 seconds respectively. After what interval of time will they toll again together?
LCM = 300/60 = 5 min
LCM = 300/60 = 5 min
What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58?
Let the side of the square plot be a ft.
a^{2} = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 * 58 = Rs. 3944.
Let the side of the square plot be a ft.
a^{2} = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 * 58 = Rs. 3944.
I. a^{2} – 13a + 42 = 0,
II. b^{2} – 15b + 56 = 0 to solve both the equations to find the values of a and b?
I. a^{2} – 13a + 42 = 0
=>(a – 6)(a – 7) = 0 => a = 6, 7
II. b^{2} – 15b + 56 = 0
=>(b – 7)(b – 8) = 0 => b = 7, 8
=>a ≤ b
I. a^{2} – 13a + 42 = 0
=>(a – 6)(a – 7) = 0 => a = 6, 7
II. b^{2} – 15b + 56 = 0
=>(b – 7)(b – 8) = 0 => b = 7, 8
=>a ≤ b
Simplify : (y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a} . y^{b} . y^{c})^{3}
(y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a} . y^{b} . y^{c})^{3}
= (y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a+b+c})^{3}
= y^{4ba+4cb+4ac} / y^{3(a+b+c)}
= y^{3(a+b+c)} / y^{3(a+b+c)} = 1.
(y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a} . y^{b} . y^{c})^{3}
= (y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a+b+c})^{3}
= y^{4ba+4cb+4ac} / y^{3(a+b+c)}
= y^{3(a+b+c)} / y^{3(a+b+c)} = 1.
A man can row 30 km downstream and 20 km upstream in 4 hours. He can row 45 km downstream and 40 km upstream in 7 hours. Find the speed of man in still water?
Let the speed of the man in still water be a kmph and let the speed of the stream be b kmph.
Now 30/(a + b) + 20/(a – b) = 4 and 45/(a + b) + 40/(a – b) = 7
Solving the equation, the speed of man in still water is 12.5 kmph.
Let the speed of the man in still water be a kmph and let the speed of the stream be b kmph.
Now 30/(a + b) + 20/(a – b) = 4 and 45/(a + b) + 40/(a – b) = 7
Solving the equation, the speed of man in still water is 12.5 kmph.
A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start in the same direction from the same point at 7.30 a.m. They shall first cross each other at?
Since A and B move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two.
Relative speed of A and B = 6 – 1 = 5 rounds per hour.
Time taken to complete one round at this speed = 1/5 hr = 12 min.
Since A and B move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two.
Relative speed of A and B = 6 – 1 = 5 rounds per hour.
Time taken to complete one round at this speed = 1/5 hr = 12 min.
Income and expenditure of a person are in the ratio 5:4. If the income of the person is Rs.18000, then find his savings?
Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs.3600.
Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs.3600.
What is the least number to be subtracted from 11, 15, 21 and 30 each so that the resultant numbers become proportional?
Let the least number to be subtracted be ‘x’, then 11 – x, 15 – x, 21 – x and 30 – x are in proportion.
<=> (11 – x):(15 – x) = (21 – x):(30 x)(21 – x)
From the options, when x = 3
=> 8 * 27 = 12 * 18 => then x = 3. => (11 – x)(30 – x) = (15 – x)(21 – x)
Let the least number to be subtracted be ‘x’, then 11 – x, 15 – x, 21 – x and 30 – x are in proportion.
<=> (11 – x):(15 – x) = (21 – x):(30 x)(21 – x)
From the options, when x = 3
=> 8 * 27 = 12 * 18 => then x = 3. => (11 – x)(30 – x) = (15 – x)(21 – x)
Two pipes can fill a tank in 18 minutes and 15 minutes. An outlet pipe can empty the tank in 45 minutes. If all the pipes are opened when the tank is empty, then how many minutes will it take to fill the tank?
Part of the filled by all the three pipes in one minute
= 1/18 + 1/15 – 1/45 = (5 + 6 – 2)/90 = 9/90 = 1/10
So, the tank becomes full in 10 minutes.
Part of the filled by all the three pipes in one minute
= 1/18 + 1/15 – 1/45 = (5 + 6 – 2)/90 = 9/90 = 1/10
So, the tank becomes full in 10 minutes.
The simple interest on Rs.12000 at a certain rate of interest in five years is Rs.7200. Find the compound interest on the same amount for five years at the same rate of interest.
R = 100 I / PT
=> R = (100 * 7200)/ (12000 * 5) = 12%
CI = P{ [1 + R /100]^{n} – 1}
= 12000 { [ 1 + 12 / 100]^{2} – 1} = Rs.3052.80
R = 100 I / PT
=> R = (100 * 7200)/ (12000 * 5) = 12%
CI = P{ [1 + R /100]^{n} – 1}
= 12000 { [ 1 + 12 / 100]^{2} – 1} = Rs.3052.80
The difference between a twodigit number and the number obtained by interchanging the digit is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1:2?
Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let the ten’s and unit’s digits be 2x and x respectively.
Then, (10 * 2x + x) – (10x + 2x) = 36
9x = 36
x = 4
Required difference = (2x + x) – (2x – x) = 2x = 8.
Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let the ten’s and unit’s digits be 2x and x respectively.
Then, (10 * 2x + x) – (10x + 2x) = 36
9x = 36
x = 4
Required difference = (2x + x) – (2x – x) = 2x = 8.
(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
=> 830 + ? * 42 – 830 * 42 = 2018
=> ? * 42 = 2018 + 830 * 42 – 830
=> ? * 42 = 42 * 48 + 830 * 42 – 830
=> ? = 48 + 830 – (830 / 42)
=> ? = 48 + 830 – 19.76
=> ? = 858.24
(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
=> 830 + ? * 42 – 830 * 42 = 2018
=> ? * 42 = 2018 + 830 * 42 – 830
=> ? * 42 = 42 * 48 + 830 * 42 – 830
=> ? = 48 + 830 – (830 / 42)
=> ? = 48 + 830 – 19.76
=> ? = 858.24
X and Y started a business with capitals Rs. 20000 and Rs. 25000. After few months Z joined them with a capital of Rs. 30000. If the share of Z in the annual profit of Rs. 50000 is Rs. 14000, then after how many months from the beginning did Z join?
Investments of X, Y and Z respectively are Rs. 20000, Rs. 25000 and Rs. 30000
Let investment period of Z be x months.
Ratio of annual investments of X, Y and Z is (20000 * 12) : (25000 * 12) : (30000 * x)
= 240 : 300 : 30x = 8 : 10 : x
The share of Z in the annual profit of Rs. 50000 is Rs. 14000.
=> [x/ (18 + x)] 50000 = 14000 => [x/ (18 + x)] 25 = 7
=> 25x = 7x + (18 * 7) => x = 7 months.
Z joined the business after (12 – 7) months. i.e., 5 months.
Investments of X, Y and Z respectively are Rs. 20000, Rs. 25000 and Rs. 30000
Let investment period of Z be x months.
Ratio of annual investments of X, Y and Z is (20000 * 12) : (25000 * 12) : (30000 * x)
= 240 : 300 : 30x = 8 : 10 : x
The share of Z in the annual profit of Rs. 50000 is Rs. 14000.
=> [x/ (18 + x)] 50000 = 14000 => [x/ (18 + x)] 25 = 7
=> 25x = 7x + (18 * 7) => x = 7 months.
Z joined the business after (12 – 7) months. i.e., 5 months.
Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in?
Net part filled in 1 hour = 1/5 + 1/6 – 1/12 = 17/60
The tank will be full in 60/17 hrs, i.e., 3 9/17 hrs.
Net part filled in 1 hour = 1/5 + 1/6 – 1/12 = 17/60
The tank will be full in 60/17 hrs, i.e., 3 9/17 hrs.
A certain number of men can do a work in 65 days working 6 hours a day. If the number of men are decreased by onefourth, then for how many hours per day should they work in order to complete the work in 40 days?
Let the number of men initially be x. we have M_{1} D_{1} H_{1}= M_{2} D_{2} H_{2}
So, x * 65 * 6 = (3x)/4 * 40 * h_{2}
=> h_{2} = (65 * 6 * 4)/(3 * 40) = 13.
Let the number of men initially be x. we have M_{1} D_{1} H_{1}= M_{2} D_{2} H_{2}
So, x * 65 * 6 = (3x)/4 * 40 * h_{2}
=> h_{2} = (65 * 6 * 4)/(3 * 40) = 13.
A gardener wants to plant trees in his garden in rows in such a way that the number of trees in each row to be the same. If there are 24 rows the number of trees in each row is 42 if there are 12 more rows find the number of trees in each row?
Required number of trees
= 24/36 * 42 = 28.
Required number of trees
= 24/36 * 42 = 28.
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A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?
Relative speed of the thief and policeman = 11 – 10 = 1 km/hr.
Distance covered in 6 minutes = 1/60 * 6 = 1/10 km = 100 m.
Distance between the thief and policeman = 200 – 100 = 100 m.
Relative speed of the thief and policeman = 11 – 10 = 1 km/hr.
Distance covered in 6 minutes = 1/60 * 6 = 1/10 km = 100 m.
Distance between the thief and policeman = 200 – 100 = 100 m.
If 10% of x = 20% of y, then x:y is equal to:
10% of x = 20% of y
10x/100 = 20y/100 => x/10 = y/5
x/y = 10/5 = 2/1
x:y = 2:1.
10% of x = 20% of y
10x/100 = 20y/100 => x/10 = y/5
x/y = 10/5 = 2/1
x:y = 2:1.
The area of a parallelogram is 128sq m and its altitude is twice the corresponding base. Then the length of the base is?
2x * x = 128 => x= 8
2x * x = 128 => x= 8
A water tank is twofifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?
Clearly, pipe B is faster than pipe A and so, the tank will be emptied.
Part to be emptied = 2/5
Part emptied by (A + B) in 1 minute = (1/6 – 1/10) = 1/15
1/15 : 2/5 :: 1 : x
x = (2/5 * 1 * 15) = 6 min.
So, the tank will be emptied in 6 min.
Clearly, pipe B is faster than pipe A and so, the tank will be emptied.
Part to be emptied = 2/5
Part emptied by (A + B) in 1 minute = (1/6 – 1/10) = 1/15
1/15 : 2/5 :: 1 : x
x = (2/5 * 1 * 15) = 6 min.
So, the tank will be emptied in 6 min.
LCM of 1/3, 5/6, 5/4, 10/7 is:
LCM of numerators = 10
HCF of denominators = 1
=> 10/1 = 10
LCM of numerators = 10
HCF of denominators = 1
=> 10/1 = 10
(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
=> 830 + ? * 42 – 830 * 42 = 2018
=> ? * 42 = 2018 + 830 * 42 – 830
=> ? * 42 = 42 * 48 + 830 * 42 – 830
=> ? = 48 + 830 – (830 / 42)
=> ? = 48 + 830 – 19.76
=> ? = 858.24
(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
=> 830 + ? * 42 – 830 * 42 = 2018
=> ? * 42 = 2018 + 830 * 42 – 830
=> ? * 42 = 42 * 48 + 830 * 42 – 830
=> ? = 48 + 830 – (830 / 42)
=> ? = 48 + 830 – 19.76
=> ? = 858.24
What is the are of an equilateral triangle of side 16 cm?
Area of an equilateral triangle = √3/4 S^{2}
If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm^{2};
Area of an equilateral triangle = √3/4 S^{2}
If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm^{2};
Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill cistern. How much time will be taken by A to fill the cistern separately?
Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
1/x + 1/(x + 6) = 1/4
x^{2} – 2x – 24 = 0
(x – 6)(x + 4) = 0 => x = 6.
Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
1/x + 1/(x + 6) = 1/4
x^{2} – 2x – 24 = 0
(x – 6)(x + 4) = 0 => x = 6.
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
The area of a square is 4096 sq cm. Find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square.
Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.
a2 = 4096 = 2^{12}
a = (2^{12})^{1/2} = 2^{6} = 64
L = 2a and b = a – 24
b : l = a – 24 : 2a = 40 : 128 = 5 : 16
Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.
a2 = 4096 = 2^{12}
a = (2^{12})^{1/2} = 2^{6} = 64
L = 2a and b = a – 24
b : l = a – 24 : 2a = 40 : 128 = 5 : 16
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
A can do a piece of work in 21 days and B in 28 days. Together they started the work and B left after 4 days. In how many days can A alone do the remaining work?
Let A worked for x days.
x/21 + 4/28 = 1 => x/21 = 6/7 => x = 18
A worked for 18 days. So, A can complete the remaining work in 18 – 4 = 14 days.
Let A worked for x days.
x/21 + 4/28 = 1 => x/21 = 6/7 => x = 18
A worked for 18 days. So, A can complete the remaining work in 18 – 4 = 14 days.
150 men consume 1050 kg of rice in 30 days. In how many days will 70 men consume 980 kg of rice?
Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day
Let us say 70 men take x days to consume 150 kg.
Quantity consumed by each item in x days = 7x/30 kg.
Quantity consumed by 70 men in x days = (7x/30)(70)kg
(7x/30)(70) = 980
x = (980 * 30)/490 => x = 60 days
Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day
Let us say 70 men take x days to consume 150 kg.
Quantity consumed by each item in x days = 7x/30 kg.
Quantity consumed by 70 men in x days = (7x/30)(70)kg
(7x/30)(70) = 980
x = (980 * 30)/490 => x = 60 days
The salary of a worker is first increased by 30% and afterwards reduced by 30%. What is net change in his salary?
(30 * 30)/100 = 9% decrease
(30 * 30)/100 = 9% decrease
A train passes a man standing on the platform. If the train is 170 meters long and its speed is 72 kmph, how much time it took in doing so?
D = 170
S = 72 * 5/18 = 20 mps
T = 170/20 = 8 ½ sec
D = 170
S = 72 * 5/18 = 20 mps
T = 170/20 = 8 ½ sec
A lends Rs. 2500 to B and a certain to C at the same time at 7% p.a. simple interest. If after 4 years, A altogether receives Rs. 1120 as interest from B and C, then the sum lent to C is?
Let the sum lent to C be Rs. x. Then,
(2500 * 7 * 4) / 100 + (x * 7 * 4) / 100 = 1120
7/25 x = (1120 – 700) => x = 1500
Let the sum lent to C be Rs. x. Then,
(2500 * 7 * 4) / 100 + (x * 7 * 4) / 100 = 1120
7/25 x = (1120 – 700) => x = 1500
The smallest ratio out of 1:1, 2:1, 1:3 and 3:1 is?
Find the fraction which has the same ratio to 2/13 that 5/34 has to 7/48.
P : 2/13 = 5/34 : 7/48
As the product of the means is equal to the product of the extremes.
P*7/48 = 2/13 * 5/34
P*7/48 = 10/442
P = 480/3094 => P = 240/1547
P : 2/13 = 5/34 : 7/48
As the product of the means is equal to the product of the extremes.
P*7/48 = 2/13 * 5/34
P*7/48 = 10/442
P = 480/3094 => P = 240/1547
A train travelled from station P to Q in 8 hours and came back from station Q to P is 6 hours. What would be the ratio of the speed of the train while traveling from station P to Q to that from station Q to P?
Since S # 1/t
S_{1} : S_{2} = 1/t_{1} : 1/t_{2} = 1/8 : 1/6 = 3 : 4
Since S # 1/t
S_{1} : S_{2} = 1/t_{1} : 1/t_{2} = 1/8 : 1/6 = 3 : 4
A certain sum amounts to Rs.1725 in 3 years and Rs.1875 in 5 years. Find the rate % per annum?
3 — 1725
5 — 1875
————–
2 — 150
N = 1 I = 75 R = ?
P = 1725 – 225 = 1500
75 = (1500*1*R)/100
R = 5%
3 — 1725
5 — 1875
————–
2 — 150
N = 1 I = 75 R = ?
P = 1725 – 225 = 1500
75 = (1500*1*R)/100
R = 5%
The volumes of two cones are in the ratio 1 : 10 and the radii of the cones are in the ratio of 1 : 2. What is the length of the wire?
The volume of the cone = (1/3)πr^{2}h
Only radius (r) and height (h) are varying.
Hence, (1/3)π may be ignored.
V_{1}/V_{2} = r_{1}^{2}h_{1}/r_{2}^{2}h_{2} => 1/10 = (1)2h_{1}/(2)^{2}h_{2}
=> h_{1}/h_{2} = 2/5
i.e. h_{1} : h_{2} = 2 : 5
The volume of the cone = (1/3)πr^{2}h
Only radius (r) and height (h) are varying.
Hence, (1/3)π may be ignored.
V_{1}/V_{2} = r_{1}^{2}h_{1}/r_{2}^{2}h_{2} => 1/10 = (1)2h_{1}/(2)^{2}h_{2}
=> h_{1}/h_{2} = 2/5
i.e. h_{1} : h_{2} = 2 : 5
The difference between the two digit number and the number obtained by interchanging its digits is 54. The difference between the digits is?
(10x + y) – (10y + x) = 54
x – y = 6
(10x + y) – (10y + x) = 54
x – y = 6
The number which exceeds 16% of it by 42 is?
Let the number be x.
Then, x – 16% of x = 42.
x – 16/100 x = 42
x = (42 * 25)/21 = 50
Let the number be x.
Then, x – 16% of x = 42.
x – 16/100 x = 42
x = (42 * 25)/21 = 50
Find the greatest number which leaves the same remainder when it divides 25, 57 and 105.
105 – 57 = 48
57 – 25 = 32
105 – 25 = 80
The H.C.F of 32, 48 and 80 is 16.
105 – 57 = 48
57 – 25 = 32
105 – 25 = 80
The H.C.F of 32, 48 and 80 is 16.
The average of first 10 even numbers is?
Sum of 10 even numbers = 10 * 11 = 110
Average = 110/10 = 11
Sum of 10 even numbers = 10 * 11 = 110
Average = 110/10 = 11
√3584 * 52.034 = ?
√3600 * 52 = ?
=> ? = 60 * 52 = 3120
√3600 * 52 = ?
=> ? = 60 * 52 = 3120
40% of Ram’s marks is equal to 20% of Rahim’s marks which percent is equal to 30% of Robert’s marks. If Robert’s marks is 80, then find the average marks of Ram and Rahim?
Given, 40% of Ram’s marks = 20% of Rahim’s marks = 30% of Robert’s marks.
Given, marks of Robert = 80
30% of 80 = 30/100 * 8 = 24
Given, 40% of Ram’s marks = 24.
=> Ram’s marks = (24 * 100)/40 = 60
Also, 20% of Rahim’s marks = 24
=> Rahim’s marks = (24 * 100)/20 = 120
Average marks of Ram and Rahim = (60 + 120)/2 = 90.
Given, 40% of Ram’s marks = 20% of Rahim’s marks = 30% of Robert’s marks.
Given, marks of Robert = 80
30% of 80 = 30/100 * 8 = 24
Given, 40% of Ram’s marks = 24.
=> Ram’s marks = (24 * 100)/40 = 60
Also, 20% of Rahim’s marks = 24
=> Rahim’s marks = (24 * 100)/20 = 120
Average marks of Ram and Rahim = (60 + 120)/2 = 90.
A sum of money becomes double itself in 8 years at simple interest. How many times will it become 10 years at the same rate?
P — 2P — 8 years
2 1/4 P — 10 years
P — 2P — 8 years
2 1/4 P — 10 years
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
³√216400 + √280 + √322 = ?
³√216400 + √280 + √322
= 60 + 16.5 + 18 = 94.5 = 95
³√216400 + √280 + √322
= 60 + 16.5 + 18 = 94.5 = 95
15% of 195 + 14% of 205 = ?
15/100 * 195 + 14/100 * 205 = 15/100 * 200 + 14/100 * 200
= 30 + 28 = 58
15/100 * 195 + 14/100 * 205 = 15/100 * 200 + 14/100 * 200
= 30 + 28 = 58
If the sum of a number and its square is 182, what is the number?
Let the number be x. Then,
x + x^{2} = 182
(x + 14)(x – 13) = 0
x = 13
Let the number be x. Then,
x + x^{2} = 182
(x + 14)(x – 13) = 0
x = 13
Two pipes function simultaneously the reservoir will be filled in 12 hours. One pipe fills reservoir 10 hours faster than the other. How many hours does the faster pipe take to fill the reservoir?
1/x + 1/(x + 10) = 1/12
x = 20
1/x + 1/(x + 10) = 1/12
x = 20
A shopkeeper buys two articles for Rs.1000 each and then sells them, making 20% profit on the first article and 20% loss on second article. Find the net profit or loss percent?
Profit on first article = 20% of 1000 = 200.
This is equal to the loss he makes on the second article. That, is he makes neither profit nor loss.
Profit on first article = 20% of 1000 = 200.
This is equal to the loss he makes on the second article. That, is he makes neither profit nor loss.
What sum of money will produce Rs.70 as simple interest in 4 years at 3 1/2 percent?
70 = (P*4*7/2)/100
P = 500
70 = (P*4*7/2)/100
P = 500
The two trains of lengths 400 m, 600 m respectively, running at same directions. The faster train can cross the slower train in 180 sec, the speed of the slower train is 48 km. then find the speed of the faster train?
Length of the two trains = 600m + 400m
Speed of the first train = X
Speed of the second train= 48 Kmph
1000/X – 48 = 180
1000/x – 48 * 5/18 = 180
50 = 9X – 120
X = 68 Kmph
Length of the two trains = 600m + 400m
Speed of the first train = X
Speed of the second train= 48 Kmph
1000/X – 48 = 180
1000/x – 48 * 5/18 = 180
50 = 9X – 120
X = 68 Kmph
What percent of 120 are 90?
(?% /100) * 120 = 90
? = 75%
(?% /100) * 120 = 90
? = 75%
The least number which when divided by 16, 18 and 21, leave the remainder 3, 5 and 8 respectively is:
2222.2 + 222.22 + 22.222 = ?
If Rs.7500 are borrowed at C.I at the rate of 4% per annum, then after 2 years the amount to be paid is?
A = 7500(26/25)^{2} = 8112
A = 7500(26/25)^{2} = 8112
men are equal to as many women as are equal to 8 boys. All of them earn Rs.90 only. Men’s wages are?
5M = xW = 8B
5M + xW + 8B —– 90 Rs.
5M + 5M + 5M —– 90 Rs.
15M —— 90 Rs. => 1M = 6Rs.
5M = xW = 8B
5M + xW + 8B —– 90 Rs.
5M + 5M + 5M —– 90 Rs.
15M —— 90 Rs. => 1M = 6Rs.
25 * 25 / 25 + 15 * 40 = ?
(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.
(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.
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.
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
What is the difference between the compound interest on Rs.12000 at 20% p.a. for one year when compounded yearly and half yearly?
When compounded annually, interest
= 12000[1 + 20/100]^{1} – 12000 = Rs.2400
When compounded semiannually, interest
= 12000[1 + 10/100]^{2} – 12000 = Rs.2520
Required difference = 2520 – 2400 = Rs.120
When compounded annually, interest
= 12000[1 + 20/100]^{1} – 12000 = Rs.2400
When compounded semiannually, interest
= 12000[1 + 10/100]^{2} – 12000 = Rs.2520
Required difference = 2520 – 2400 = Rs.120
60 + 5 * 12 / (180/3) = ?
60 + 5 * 12 / (180/3) = 60 + 5 * 12 / (60)
= 60 + (5 * 12)/60 = 60 + 1 = 61.
60 + 5 * 12 / (180/3) = 60 + 5 * 12 / (60)
= 60 + (5 * 12)/60 = 60 + 1 = 61.
Find the one which does not belong to that group ?
Part, Cart, Dart and Mart are rhyming words, but Trap does not sound similarly.