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RBI Assistant Mains Mock Test, RBI Assistant mains online test series 2. RBI Exam Online Test 2021, Take CAknowledge RBI Assistant Mains Free Quiz Exam 2021. RBI Assistant Mains Exam Free Online Quiz 2021, RBI Assistant Mains Full Online Mock Test Series 2nd in English. To Analyse your preparation, one should attempt the test series on a regular basis to score more in the examination. Start FREE mock test and get top rank among all applicants whoever are participating in the exam. RBI Assistant Mains Question and Answers in English and Hindi Series 2. Here we are providing RBI Assistant Mains Full Mock Test Paper in English. RBI Assistant Mains Mock Test Series 2nd 2021. Now Test your self for RBI Assistant Mains Exam by using below quiz…
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[10 + {4 * ({2/3 + 1/4} * √144/121 + 23) ÷ 12 + 5} – 3] = ?
[10 + { 4 * (2/3 + 1/4 * √144/121 + 23) ÷ 12 + 5} – 3] = ? [10 + { 4 * (11/12 * 12/11 + 23) ÷ 12 + 5 } – 3] = ?
= [10 + {4 * 24 ÷ 12 + 5} – 3] = ?
= [10 + 13 – 3] = ? = > 20
[10 + { 4 * (2/3 + 1/4 * √144/121 + 23) ÷ 12 + 5} – 3] = ? [10 + { 4 * (11/12 * 12/11 + 23) ÷ 12 + 5 } – 3] = ?
= [10 + {4 * 24 ÷ 12 + 5} – 3] = ?
= [10 + 13 – 3] = ? = > 20
Two tests had the same maximum mark. The pass percentages in the first and the second test were 40% and 45% respectively. A candidate scored 216 marks in the second test and failed by 36 marks in that test. Find the pass mark in the first test?
Let the maximum mark in each test be M.
The candidate failed by 36 marks in the second test.
pass mark in the second test = 216 + 36 = 252
45/100 M = 252
Pass mark in the first test = 40/100 M = 40/45 * 252 = 224.
Let the maximum mark in each test be M.
The candidate failed by 36 marks in the second test.
pass mark in the second test = 216 + 36 = 252
45/100 M = 252
Pass mark in the first test = 40/100 M = 40/45 * 252 = 224.
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.
2222.2 + 222.22 + 22.222 = ?
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.
[3/7 of {21 * (33.33% of (27 + 3) ÷ 5) – 7} – 8] = ?
[3/7 of {21 * (33.33% of (27 + 3) ÷ 5) – 7} – 8] = ?
= [3/7 of {21 * (1/3 * 30 ÷ 5) – 7} – 8] = ?
= 3/7 of [{21 * 2 – 7} – 8] = ? => {3/7 * 35 – 8} = ? => 7
[3/7 of {21 * (33.33% of (27 + 3) ÷ 5) – 7} – 8] = ?
= [3/7 of {21 * (1/3 * 30 ÷ 5) – 7} – 8] = ?
= 3/7 of [{21 * 2 – 7} – 8] = ? => {3/7 * 35 – 8} = ? => 7
What percentage of numbers from 1 to 70 have squares that end in the digit 1?
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such numbers = 14.
Required percentage = (14/70 * 100) = 20%
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such numbers = 14.
Required percentage = (14/70 * 100) = 20%
(5 – 7)^{2} * (6 – 9)^{2} / (2^{3})^{2} = ?
(5 – 7)^{2} * (6 – 9)^{2} / (2^{3})^{2} = (2)^{2} * (3)^{2} * 1/(2^{3})^{2}
= 4 * 9 * 64 = 2304.
(5 – 7)^{2} * (6 – 9)^{2} / (2^{3})^{2} = (2)^{2} * (3)^{2} * 1/(2^{3})^{2}
= 4 * 9 * 64 = 2304.
Six years ago, the ratio of ages of Kunal and Sagar was 6:5. Four years hence, the ratio of their ages will be 11:10. What is Sagar’s age at present?
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then, [(6x + 6) + 4] / [(5x + 6) + 4] = 11/10
10(6x + 10) = 11(5x + 10) => x = 2
Sagar’s present age = (5x + 6) = 16 years.
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then, [(6x + 6) + 4] / [(5x + 6) + 4] = 11/10
10(6x + 10) = 11(5x + 10) => x = 2
Sagar’s present age = (5x + 6) = 16 years.
Mudit’s age 18 years hence will be thrice his age four years ago. Find Mudit’s present age?
Let Mudit’s present age be ‘m’ years.
m + 18 = 3(m – 4)
=> 2m = 30 => m = 15 years.
Let Mudit’s present age be ‘m’ years.
m + 18 = 3(m – 4)
=> 2m = 30 => m = 15 years.
The least number of four digits which is divisible by 4, 6, 8 and 10 is?
LCM = 120
120) 1000 (8
960
——
40
1000 + 120 – 40 = 1080
LCM = 120
120) 1000 (8
960
——
40
1000 + 120 – 40 = 1080
The ratio of investments of two partners P and Q is 7:5 and the ratio of their profits is 7:10. If P invested the money for 5 months, find for how much time did Q invest the money?
7*5: 5*x = 7:10
x = 10
7*5: 5*x = 7:10
x = 10
There are some rabbits and peacocks in a zoo. The total number of their heads is 60 and total number of their legs is 192. Find the number of total rabbits?
Let the number of rabbits and peacocks be ‘r’ and ‘p’ respectively. As each animal has only one head, so r + p = 60 — (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 — (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36.
Let the number of rabbits and peacocks be ‘r’ and ‘p’ respectively. As each animal has only one head, so r + p = 60 — (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 — (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36.
A, B, C together started a business. A invested Rs.6000 for 5 months B invested Rs.3600 for 6 months and C Rs.7500 for 3 months. If they get a total profit of Rs.7410. Find the share of A?
60*5:36*6:75*3
100: 72: 75
100/247 * 7410 = 3000
60*5:36*6:75*3
100: 72: 75
100/247 * 7410 = 3000
The amount of water (in ml) that should be added to reduce 9 ml. Lotion, containing 50% alcohol, to a lotion containing 30% alcohol, is?
4.5 4.5
30% 70%
30% —– 4.5
70% ——? => 10.5 – 4.5 = 6 ml
4.5 4.5
30% 70%
30% —– 4.5
70% ——? => 10.5 – 4.5 = 6 ml
A dishonest dealer professes to sell goods at the cost price but uses a false weight and gains 25%. Find his false weight age?
25 = E/(1000 – E) * 100
1000 – E = 4E
1000 = 5E => E = 200
1000 – 200 = 800
25 = E/(1000 – E) * 100
1000 – E = 4E
1000 = 5E => E = 200
1000 – 200 = 800
Find the one which does not belong to that group ?
Boxing, Chess, Wrestling and Squash are individual events, while Baseball is a team event.
Boxing, Chess, Wrestling and Squash are individual events, while Baseball is a team event.
A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width. Find the least number of square tiles of equal size required to cover the entire floor of the room?
Length = 6 m 24 cm = 624 cm
Width = 4 m 32 cm = 432 cm
HCF of 624 and 432 = 48
Number of square tiles required = (624 * 432)/(48 * 48) = 13 * 9 = 117
Length = 6 m 24 cm = 624 cm
Width = 4 m 32 cm = 432 cm
HCF of 624 and 432 = 48
Number of square tiles required = (624 * 432)/(48 * 48) = 13 * 9 = 117
Rs. 1300 is divided into three parts A, B and C. How much A is more than C if their ratio is 1/2:1/3:1/4?
1/2:1/3:1/4 = 6:4:3
3/13*1300 = 300
1/2:1/3:1/4 = 6:4:3
3/13*1300 = 300
[8 + {(21 * √9/441 – 1 /5 of 60% – 1/5) * 625 + 7} ÷ 4] = ?
[8 + {(21 * √9/441 – 1 /5 of 60% – 1/5) * 625 + 7} ÷ 4] = ?
=[8 + [(2/5 * 3/5 – 1/5) * 625 + 7} ÷ 4] = ?
=[8 + {(6 – 5 /25) * 625 + 7} ÷ 4 = ?
=[8 + {32} ÷ 4] = ? = 16
[8 + {(21 * √9/441 – 1 /5 of 60% – 1/5) * 625 + 7} ÷ 4] = ?
=[8 + [(2/5 * 3/5 – 1/5) * 625 + 7} ÷ 4] = ?
=[8 + {(6 – 5 /25) * 625 + 7} ÷ 4 = ?
=[8 + {32} ÷ 4] = ? = 16
Find the one which does not belong to that group ?
11 = 1 1^{3}, 28 = 2 2^{3}, 327 = 3 3^{3}, 416 = 4 4^{2} and 5125 = 5 5^{3}
Except 416, other numbers follow similar pattern.
11 = 1 1^{3}, 28 = 2 2^{3}, 327 = 3 3^{3}, 416 = 4 4^{2} and 5125 = 5 5^{3}
Except 416, other numbers follow similar pattern.
A certain sum of money is invested for one year at a certain rate of simple interest. If the rate of interest is 3% higher, then the invest earned will be 25% more than the interest earned earlier. What is the earlier rate of interest?
If the interest earned is 25% more than the earlier interest then the rate of interest also should be 25% higher than the earlier rate.
Let the earlier rate of interest be x%.
Now it will be (x + 3)%
% increase = (x + 3) – x/x * 100 = 25
=> x = 12
If the interest earned is 25% more than the earlier interest then the rate of interest also should be 25% higher than the earlier rate.
Let the earlier rate of interest be x%.
Now it will be (x + 3)%
% increase = (x + 3) – x/x * 100 = 25
=> x = 12
The area of a triangle will be when a = 1m, b = 2m, c = 3m, a, b, c being lengths of respective sides.
S = (1 + 2 + 3)/2 = 3
=> No triangle exists
S = (1 + 2 + 3)/2 = 3
=> No triangle exists
Find the area of a rhombus whose side is 25 cm and one of the diagonals is 30 cm?
Consider the rhombus ABCD. Let the diagonals intersect at E. Since diagonals
bisect at right angles in a rhombus.
BE^{2} + AE^{2} = AB^{2}
25^{2} = 15^{2} + AE^{2} AE = √(625 – 225) = √400 = 20,
AC = 20 + 20 = 40 cm.
Area of a rhombus = 1/2 * d_{1}d_{2}
= 1/2 * 40 * 30 = 600 sq.cm.
Consider the rhombus ABCD. Let the diagonals intersect at E. Since diagonals
bisect at right angles in a rhombus.
BE^{2} + AE^{2} = AB^{2}
25^{2} = 15^{2} + AE^{2} AE = √(625 – 225) = √400 = 20,
AC = 20 + 20 = 40 cm.
Area of a rhombus = 1/2 * d_{1}d_{2}
= 1/2 * 40 * 30 = 600 sq.cm.
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 G.C.D of 1.08, 0.36 and 0.9 is
Given numbers are 1.08, 0.36 and 0.90.
H.C.F of 108, 36 and 90 is 18.
H.C.F of a given numbers = 0.18
Given numbers are 1.08, 0.36 and 0.90.
H.C.F of 108, 36 and 90 is 18.
H.C.F of a given numbers = 0.18
Find the one which does not belong to that group ?
Except Japanian, all others are appropriate usage of citizenship.
Except Japanian, all others are appropriate usage of citizenship.
If the perimeter of a rectangular garden is 600 m, its length when its breadth is 100 m is?
2(l + 100) = 600 => l = 200 m
2(l + 100) = 600 => l = 200 m
A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that exactly three bulbs are good.
Required probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63
Required probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63
The curved surface of a sphere is 64 π cm^{2}. Find its radius?
4 πr^{2 }= 64 => r = 4
4 πr^{2 }= 64 => r = 4
P is three times as fast as Q and working together, they can complete a work in 12 days. In how many days can Q alone complete the work?
P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 12 days, which means Q can do the work in 48 days.
Hence, P can do the work in 16 days.
P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 12 days, which means Q can do the work in 48 days.
Hence, P can do the work in 16 days.
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
A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is?
S.I. for 3 years = (12005 – 9800) = Rs. 2205
S.I. for 5 years = Rs. 2205/3 * 5 = Rs. 3675.
Principal = (9800 – 3675) = Rs. 6125
Hence, rate = (100 * 3675) / (6125 * 5) = 12%
S.I. for 3 years = (12005 – 9800) = Rs. 2205
S.I. for 5 years = Rs. 2205/3 * 5 = Rs. 3675.
Principal = (9800 – 3675) = Rs. 6125
Hence, rate = (100 * 3675) / (6125 * 5) = 12%
The area of a square is equal to five times the area of a rectangle of dimensions 125 cm * 64 cm. What is the perimeter of the square?
Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm.
Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm.
Walking 7/6 of his usual rate, a boy reaches his school 4 min early. Find his usual time to reach the school?
Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? → 28 m
Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? → 28 m
In what ratio should two varieties of sugar of Rs.18 per kg and Rs.24 kg be mixed together to get a mixture whose cost is Rs.20 per kg?
(50 – ?/29)% of 4200 = 3√196
(50 – ?/29)% of 4200 = 3√196
(50 – ?/29)% of 4200 = 3√196
Sachin is younger than Rahul by 4 years. If their ages are in the respective ratio of 7:9, how old is Sachin?
Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
(x – 7)/x = 7/9
2x = 63 => x = 31.5
Hence, Sachin’s age = (x – 7) = 24.5 years.
Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
(x – 7)/x = 7/9
2x = 63 => x = 31.5
Hence, Sachin’s age = (x – 7) = 24.5 years.
Mohit sold an article for Rs. 18000. Had he offered a discount of 10% on the selling price, he would have earned a profit of 8%. What is the cost price of the article?
Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 – 10/100(18000) = 18000 – 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000
Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 – 10/100(18000) = 18000 – 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000
The price of an article has been reduced by 25%. In order to restore the original price the new price must be increased by?
100
75
——
75 —— 25
100 —— ? => 33 1/3%
100
75
——
75 —— 25
100 —— ? => 33 1/3%
Two numbers are in the ratio 3:5. If 9 be subtracted from each, they are in the ratio of 9:17. The first number is:
(3x9):(5x9) = 9:17
x = 12 => 3x = 36
(3x9):(5x9) = 9:17
x = 12 => 3x = 36
The sale price of an article including the sales tax is Rs. 616. The rate of sales tax is 10%. If the shopkeeper has made a profit of 12%, then the cost price of the article is:
110% of S.P. = 616
S.P. = (616 * 100)/110 = Rs. 560
C.P = (110 * 560)/112 = Rs. 500
110% of S.P. = 616
S.P. = (616 * 100)/110 = Rs. 560
C.P = (110 * 560)/112 = Rs. 500
Two persons A and B take a field on rent. A puts on it 21 horses for 3 months and 15 cows for 2 months; B puts 15 cows for 6months and 40 sheep for 7 1/2 months. If one day, 3 horses eat as much as 5 cows and 6 cows as much as 10 sheep, what part of the rent should A pay?
3h = 5c
6c = 10s
A = 21h*3 + 15c*2
= 63h + 30c
= 105c + 30c = 135c
B = 15c*6 + 40s*7 1/2
= 90c + 300s
= 90c + 180c = 270c
A:B = 135:270
27:52
A = 27/79 = 1/3
3h = 5c
6c = 10s
A = 21h*3 + 15c*2
= 63h + 30c
= 105c + 30c = 135c
B = 15c*6 + 40s*7 1/2
= 90c + 300s
= 90c + 180c = 270c
A:B = 135:270
27:52
A = 27/79 = 1/3
What is the difference between the largest and the smallest number written with 7, 7, 0, 7?
7770
7077
————
693
7770
7077
————
693
What distance will be covered by a bus moving at 72 kmph in 30 seconds?
72 kmph = 72 * 5/18 = 20 mps
D = Speed * time = 20 * 30 = 600 m.
72 kmph = 72 * 5/18 = 20 mps
D = Speed * time = 20 * 30 = 600 m.
A man swims downstream 30 km and upstream 18 km taking 3 hours each time, what is the speed of the man in still water?
30 — 3 DS = 10
? — 1
18 — 3 US = 6
? — 1 M = ?
M = (10 + 6)/2 = 8
30 — 3 DS = 10
? — 1
18 — 3 US = 6
? — 1 M = ?
M = (10 + 6)/2 = 8
A, B and C play a cricket match. The ratio of the runs scored by them in the match is A:B = 2:3 and B:C = 2:5. If the total runs scored by all of them are 75, the runs scored by B are?
A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18
A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18
Thirty men can do a work in 24 days. In how many days can 20 men can do the work, given that the time spent per day is increased by onethird of the previous time?
Let the number of hours working per day initially be x. we have M_{1} D_{1} H_{1}= M_{2} D_{2} H_{2}
30 * 24 * x = 20 * d_{2} * (4x)/3 => d_{2} = (30 * 24 * 3)/(24 * 4) = 27 days.
Let the number of hours working per day initially be x. we have M_{1} D_{1} H_{1}= M_{2} D_{2} H_{2}
30 * 24 * x = 20 * d_{2} * (4x)/3 => d_{2} = (30 * 24 * 3)/(24 * 4) = 27 days.
What is the difference between the largest number and the least number written with the figures 3, 4, 7, 0, 3?
74330 Largest
30347 Smallest
————
43983
74330 Largest
30347 Smallest
————
43983
A sum of money is to be distributed among A, B, C, D in the proportion of 5:2:4:3. If C gets Rs. 1000 more than D, what is B’s share?
Let the shares of A, B, C and D be 5x, 2x, 4x and 3x Rs. respectively.
Then, 4x – 3x = 1000 => x = 1000.
B’s share = Rs. 2x = 2 * 1000 = Rs. 2000.
Let the shares of A, B, C and D be 5x, 2x, 4x and 3x Rs. respectively.
Then, 4x – 3x = 1000 => x = 1000.
B’s share = Rs. 2x = 2 * 1000 = Rs. 2000.
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Indian Bank PO Prelims Online Test Series 5th, Indian Bank Mock test Series 5. Indian Bank PO Pre Free Mock Test Exam 2021. CAknowledge provide free Indian Bank PO Pre Exam Online Quiz 2021. Indian Bank every year invites application from the young and bright candidates for the post of JMG I as Probationary officers. Indian Bank conducts online recruitment to admission to postgraduate diploma in banking & finance course offered through Manipal global education services 2021. Here we are providing Indian Bank PO Pre. Full Mock Test Paper in English. Indian Bank PO Pre. Mock Test Series 5th 2021. Now Test your self for Indian Bank PO Pre. Exam by using below quiz…
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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
Thrice the square of a natural number decreased by 4 times the number is equal to 50 more than the number. The number is:
Let the number be x. Then,
3x^{2} – 4x = x + 50
3x^{2} – 5x – 50 = 0
(3x + 10)(x – 5) = 0
x = 5
Let the number be x. Then,
3x^{2} – 4x = x + 50
3x^{2} – 5x – 50 = 0
(3x + 10)(x – 5) = 0
x = 5
What is the ratio between perimeters of two squares one having 3 times the diagonal then the other?
d = 3d d = d
a√2 = 3d a√2 = d
a = 3d/√2 a = d/√2 => 3: 1
d = 3d d = d
a√2 = 3d a√2 = d
a = 3d/√2 a = d/√2 => 3: 1
A take twice as much time as B or thrice as much time to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in?
Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
6/x = 1/2 => x = 12
So, B takes 6 hours to finish the work.
Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
6/x = 1/2 => x = 12
So, B takes 6 hours to finish the work.
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%
(18^{2} – 9^{2} * 3) / (675 * ?) = 4
(18^{2} – 9^{2} * 3) / (675 * ?) = 4
=> 27 * ? = 324 – 243
=> 27 * ? = 27(12 – 9) => ? = 3
(18^{2} – 9^{2} * 3) / (675 * ?) = 4
=> 27 * ? = 324 – 243
=> 27 * ? = 27(12 – 9) => ? = 3
Calculate the number of bricks, each measuring 25 cm * 15 cm * 8 cm required to construct a wall of dimensions 10 m * 4 m * 5 m when 10% of its volume is occupied by mortar?
10 * 4/100 * 5 * 90/100 = 25/100 * 15/100 * 8/100 * x
10 * 20 * 90 = 15 * 2 * x => x = 6000
10 * 4/100 * 5 * 90/100 = 25/100 * 15/100 * 8/100 * x
10 * 20 * 90 = 15 * 2 * x => x = 6000
A 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec. What is the length of the platform?
Speed = 300/18 = 50/3 m/sec.
Let the length of the platform be x meters.
Then, (x + 300)/39 = 50/3
3x + 900 = 1950 => x = 350 m.
Speed = 300/18 = 50/3 m/sec.
Let the length of the platform be x meters.
Then, (x + 300)/39 = 50/3
3x + 900 = 1950 => x = 350 m.
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.
In an exam, a candidate secured 504 marks our of the maximum mark of M. If the maximum mark M is converted into 800 marks, he would have secured 384 marks. What is the value of M?
504/M = 384/800
(504 * 800) / 384 = M
M = 1050
504/M = 384/800
(504 * 800) / 384 = M
M = 1050
A sum of money is borrowed and paid back in two annual installments of Rs. 882 each allowing 5% C.I. The sum borrowed was?
Principal = (P.W. of Rs. 882 due 1 year hence) + (P.W. of Rs. 882 due 2 years hence)
= [882/(1 + 5/100) + 882/(1 + 5/100)^{2}]
= (882 * 20)/21 + (882 * 400)/441 = Rs. 1640.
Principal = (P.W. of Rs. 882 due 1 year hence) + (P.W. of Rs. 882 due 2 years hence)
= [882/(1 + 5/100) + 882/(1 + 5/100)^{2}]
= (882 * 20)/21 + (882 * 400)/441 = Rs. 1640.
What sum of money put at C.I amounts in 2 years to Rs.8820 and in 3 years to Rs.9261?
8820 — 441
100 — ? => 5%
x *105/100 * 105/100 = 8820
x*1.1025=8820
x=8820/1.1025 => 8000
8820 — 441
100 — ? => 5%
x *105/100 * 105/100 = 8820
x*1.1025=8820
x=8820/1.1025 => 8000
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
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 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.
Find the one which does not belong to that group ?
27, 36, 72 and 45 are divisible by 9, but not 30.
27, 36, 72 and 45 are divisible by 9, but not 30.
How much time will take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
Time = (100 * 81) / (450 * 4.5) = 4 years
Time = (100 * 81) / (450 * 4.5) = 4 years
Find the length of the wire required to go 15 times round a square field containing 69696 m^{2}.
a^{2} = 69696 => a = 264
4a = 1056
1056 * 15 = 15840
a^{2} = 69696 => a = 264
4a = 1056
1056 * 15 = 15840
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 5 seniors and 5 juniors?
Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇
Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇
Find the roots of quadratic equation: x^{2} + x – 42 = 0?
x^{2} + 7x – 6x + 42 = 0
x(x + 7) – 6(x + 7) = 0
(x + 7)(x – 6) = 0 => x = 7, 6
x^{2} + 7x – 6x + 42 = 0
x(x + 7) – 6(x + 7) = 0
(x + 7)(x – 6) = 0 => x = 7, 6
In a covering a certain distance, the speeds of A and B are in the ratio of 3:4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is?
Ratio of speeds = 3:4
Ratio of times taken = 4:3
Suppose A takes 4x hrs and B takes 3x hrs to reach the destination.
Then, 4x – 3x = 30/60 => x = 1/2
Time taken by A = 4x hrs = 4 * 1/2 = 2 hrs.
Ratio of speeds = 3:4
Ratio of times taken = 4:3
Suppose A takes 4x hrs and B takes 3x hrs to reach the destination.
Then, 4x – 3x = 30/60 => x = 1/2
Time taken by A = 4x hrs = 4 * 1/2 = 2 hrs.
A can do a piece of work in 15 days and B in 20 days. They began the work together but 5 days before the completion of the work, A leaves. The work was completed in?
(x – 5)/15 + x/20 = 1
x = 11 3/7 days
(x – 5)/15 + x/20 = 1
x = 11 3/7 days
What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12% p.a.?
Amount = [25000 * (1 + 12/100)^{3}]
= 25000 * 28/25 * 28/25 * 28/25 = Rs. 35123.20
C.I. = (35123.20 – 25000) = Rs. 10123.20
Amount = [25000 * (1 + 12/100)^{3}]
= 25000 * 28/25 * 28/25 * 28/25 = Rs. 35123.20
C.I. = (35123.20 – 25000) = Rs. 10123.20
In a mixture of milk and water, the proportion of milk by weight was 80%. If, in a 180 gm mixture, 36 gms of pure milk is added, what would be the percentage of milk in the mixture formed?
Percentage of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100% = (144 + 36)/216 * 100% = 5/6 * 100% = 83.33%.
Percentage of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100% = (144 + 36)/216 * 100% = 5/6 * 100% = 83.33%.
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 dishonest dealer professes to sell his goods at Cost Price but still gets 20% profit by using a false weight. What weight does he substitute for a kilogram?
If the cost price is Rs.100, then to get a profit of 20%, the selling price should be Rs.120.
If 120kg are to be sold, and the dealer gives only 100kg, to get a profit of 20%.
How many grams he has to give instead of one kilogram(1000 gm).
120 gm —— 100 gm
1000 gm —— ?
(1000 * 100)/120 = 2500/3 = 833 1/3 grams.
If the cost price is Rs.100, then to get a profit of 20%, the selling price should be Rs.120.
If 120kg are to be sold, and the dealer gives only 100kg, to get a profit of 20%.
How many grams he has to give instead of one kilogram(1000 gm).
120 gm —— 100 gm
1000 gm —— ?
(1000 * 100)/120 = 2500/3 = 833 1/3 grams.
A father said to his son, “I was as old as you are at present at the time of your birth.” If the father’s age is 38 years now, the son’s age five years back was:
Let the son’s present age be x years.
Then, (38 – x) = x
2x = 38 => x = 19
Son’s age 5 years back = (19 – 5) = 14 years.
Let the son’s present age be x years.
Then, (38 – x) = x
2x = 38 => x = 19
Son’s age 5 years back = (19 – 5) = 14 years.
The radius of the base of cone is 3 cm and height is 4 cm. Find the volume of the cone?
1/3 * π * 3 * 3 * 4 = 12 π
1/3 * π * 3 * 3 * 4 = 12 π
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 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
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 average mark of the students of a class in a particular exam is 80. If 5 students whose average mark in that exam is 40 are excluded, the average mark of the remaining will be 90. Find the number of students who wrote the exam.
Let the number of students who wrote the exam be x.
Total marks of students = 80 x.
Total marks of (x – 5) students = 90(x – 5)
80x – (5 * 40) = 90(x – 5)
250 = 10x => x = 25
Let the number of students who wrote the exam be x.
Total marks of students = 80 x.
Total marks of (x – 5) students = 90(x – 5)
80x – (5 * 40) = 90(x – 5)
250 = 10x => x = 25
A 270 m long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 sec. What is the length of the other train?
Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 270)/9 = 500/9 => x = 230.
Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 270)/9 = 500/9 => x = 230.
The first three terms of a proportion are 3, 9 and 12. The fourth term is?
(9*12)/3 = 36
(9*12)/3 = 36
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
10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
1 man’s 1 day work = 1/100
(10 men + 15 women)’s 1 day work = 1/6
15 women’s 1 day work = (1/6 – 10/100) = 1/15
1 woman’s 1 day work = 1/225
1 woman alone can complete the work in 225 days.
1 man’s 1 day work = 1/100
(10 men + 15 women)’s 1 day work = 1/6
15 women’s 1 day work = (1/6 – 10/100) = 1/15
1 woman’s 1 day work = 1/225
1 woman alone can complete the work in 225 days.
A candidate who gets 30% of the marks fails by 50 marks. But another candidate who gets 45% marks gets 25 marks more than necessary for passing. Find the number of marks for passing?
30% ———— 50
45% ———— 25
———————
15% ———— 75
30% ————– ?
150 + 50 = 200 Marks
30% ———— 50
45% ———— 25
———————
15% ———— 75
30% ————– ?
150 + 50 = 200 Marks
The incomes of two persons A and B are in the ratio 3:4. If each saves Rs.100 per month, the ratio of their expenditures is 1:2 . Find their incomes?
The incomes of A and B be 3P and 4P.
Expenditures = Income – Savings
(3P – 100) and (4P – 100)
The ratio of their expenditure = 1:2
(3P – 100):(4P – 100) = 1:2
2P = 100 => P = 50
Their incomes = 150, 200
The incomes of A and B be 3P and 4P.
Expenditures = Income – Savings
(3P – 100) and (4P – 100)
The ratio of their expenditure = 1:2
(3P – 100):(4P – 100) = 1:2
2P = 100 => P = 50
Their incomes = 150, 200
A, B and C rents a pasture for Rs.870. A put in 12 horses for 8 months, B 16 horses for 9 months and 18 horses for 6 months. How much should C pay?
12*8 :16*9 = 18*6
8: 12: 9
9/29 * 870 = 270
12*8 :16*9 = 18*6
8: 12: 9
9/29 * 870 = 270
1296 ÷ (24 * 0.75) = ?
1296 ÷ 18 = ?
=> ? = 72
1296 ÷ 18 = ?
=> ? = 72
On dividing number by 5, we get 3 as remainder. What will be the remainder when the square of this number is divided by 5?
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
If a and b are the roots of the equation x^{2} – 9x + 20 = 0, find the value of a^{2} + b^{2} + ab?
a^{2} + b^{2} + ab = a^{2} + b^{2} + 2ab – ab
i.e., (a + b)^{2} – ab
from x^{2} – 9x + 20 = 0, we have
a + b = 9 and ab = 20. Hence the value of required expression (9)^{2} – 20 = 61.
a^{2} + b^{2} + ab = a^{2} + b^{2} + 2ab – ab
i.e., (a + b)^{2} – ab
from x^{2} – 9x + 20 = 0, we have
a + b = 9 and ab = 20. Hence the value of required expression (9)^{2} – 20 = 61.
When 73^{2} is subtracted form the square of a number, the answer that is obtained is 5280. What is the number?
Let the number be x
x^{2} – 73^{2} = 5280
x^{2} = 5280 + (70+3)^{2} = 5280 + 4900 + 420 + 9 = 10609
= 10000 + 2(100)(3) + 3^{2} = (100 + 3)^{2}
x = 100 + 3 = 103.
Let the number be x
x^{2} – 73^{2} = 5280
x^{2} = 5280 + (70+3)^{2} = 5280 + 4900 + 420 + 9 = 10609
= 10000 + 2(100)(3) + 3^{2} = (100 + 3)^{2}
x = 100 + 3 = 103.
The principal that amounts to Rs. 4913 in 3 years at 6 1/4 % per annum C.I. compounded annually, is?
Principal = [4913 / (1 + 25/(4 * 100))^{3}]
= 4913 * 16/17 * 16/17 * 16/17 = Rs. 4096.
Principal = [4913 / (1 + 25/(4 * 100))^{3}]
= 4913 * 16/17 * 16/17 * 16/17 = Rs. 4096.
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
Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x^{2} + 8x + 4 = 0?
a/b + b/a = (a^{2} + b^{2})/ab = (a^{2} + b^{2} + a + b)/ab
= [(a + b)^{2} – 2ab]/ab
a + b = 8/1 = 8
ab = 4/1 = 4
Hence a/b + b/a = [(8)^{2} – 2(4)]/4 = 56/4 = 14.
a/b + b/a = (a^{2} + b^{2})/ab = (a^{2} + b^{2} + a + b)/ab
= [(a + b)^{2} – 2ab]/ab
a + b = 8/1 = 8
ab = 4/1 = 4
Hence a/b + b/a = [(8)^{2} – 2(4)]/4 = 56/4 = 14.
A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days?
Ratio of times taken by A and B = 100:130 = 10:13
Suppose B takes x days to do the work.
x = (23 * 13)/10 = 299/10
A’s 1 day work = 1/23; B’s 1 day work = 10/299
(A + B)’s 1 day work = (1/23 + 10/299) = 1/13
A and B together can complete the job in 13 days.
Ratio of times taken by A and B = 100:130 = 10:13
Suppose B takes x days to do the work.
x = (23 * 13)/10 = 299/10
A’s 1 day work = 1/23; B’s 1 day work = 10/299
(A + B)’s 1 day work = (1/23 + 10/299) = 1/13
A and B together can complete the job in 13 days.
1600 men have provisions for 28 days in the temple. If after 4 days, 400 men leave the temple, how long will the food last now?
1600 — 28 days
1600 — 24
1200 — ?
1600*24 = 1200*x
x = 32 days
1600 — 28 days
1600 — 24
1200 — ?
1600*24 = 1200*x
x = 32 days
How long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 170 m in length?
D = 110 + 170 = 280 m
S = 60 * 5/18 = 50/3
T = 280 * 3/50 = 16.8 sec
D = 110 + 170 = 280 m
S = 60 * 5/18 = 50/3
T = 280 * 3/50 = 16.8 sec
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Mahesh marks an article 15% above the cost price of Rs. 540. What must be his discount percentage if he sells it at Rs. 496.80?
CP = Rs. 540, MP = 540 + 15% of 540 = Rs. 621
SP = Rs. 496.80, Discount = 621 – 496.80 = 124.20
Discount % = 124.2/621 * 100 = 20%
CP = Rs. 540, MP = 540 + 15% of 540 = Rs. 621
SP = Rs. 496.80, Discount = 621 – 496.80 = 124.20
Discount % = 124.2/621 * 100 = 20%
The side of a square is increased by 25% then how much % does its area increases?
a = 100 a^{2} = 10000
a = 125 a^{2} = 15625
—————
10000 ——— 5625
100 ——? => 56.25 %
a = 100 a^{2} = 10000
a = 125 a^{2} = 15625
—————
10000 ——— 5625
100 ——? => 56.25 %
Sam invested Rs. 15000 @ 10% per annum for one year. If the interest is compounded halfyearly, then the amount received by Sam at the end of the year will be?
P = Rs. 15000; R = 10% p.a. = 5% per halfyear; T = 1 year = 2 halfyear
Amount = [15000 * (1 + 5/100)^{2}]
= (15000 * 21/20 * 21/20) = Rs. 16537.50
P = Rs. 15000; R = 10% p.a. = 5% per halfyear; T = 1 year = 2 halfyear
Amount = [15000 * (1 + 5/100)^{2}]
= (15000 * 21/20 * 21/20) = Rs. 16537.50
A can do a piece of work in 4 days. B can do it in 5 days. With the assistance of C they completed the work in 2 days. Find in how many days can C alone do it?
C = 1/2 – 1/4 – 1/5 = 1/20 => 20 days
C = 1/2 – 1/4 – 1/5 = 1/20 => 20 days
A man buys an article and sells it at a profit of 20%. If he had bought it at 20% less and sold it for Rs.75 less, he could have gained 25%. What is the cost price?
CP_{1} = 100 SP_{1} = 120
CP_{2} = 80 SP_{2} = 80 * (125/100) = 100
20 —– 100
75 —– ? => 375
CP_{1} = 100 SP_{1} = 120
CP_{2} = 80 SP_{2} = 80 * (125/100) = 100
20 —– 100
75 —– ? => 375
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.
The least number which when add 3 to it, completely divided 8, 12 and 18 is?
LCM = 72
72 – 3 = 69
LCM = 72
72 – 3 = 69
Twelve men can complete a piece of work in 32 days. The same work can be completed by 16 women in 36 days and by 48 boys in 16 days. Find the time taken by one man, one woman and one boy working together to complete the work?
12 men take 32 days to complete the work.
One man will take (12 * 32) days to complete it.
Similarly one woman will take (16 * 36) days to complete it and one boy will take (48 * 16) days to complete it.
One man, one woman and one boy working together will complete = 1/(12 * 32) + 1/(16 * 36) + 1/(48 * 16)
= 1/(4 * 3 * 16 * 2) + 1/(16 * 4 * 9) + 1/(16 * 3 * 4 * 4)
= 1/64(1/6 + 1/9 + 1/12) = 1/64 * 13/36 of the work in a day.
They will take (64 * 36)/13 days to complete the work working together.
12 men take 32 days to complete the work.
One man will take (12 * 32) days to complete it.
Similarly one woman will take (16 * 36) days to complete it and one boy will take (48 * 16) days to complete it.
One man, one woman and one boy working together will complete = 1/(12 * 32) + 1/(16 * 36) + 1/(48 * 16)
= 1/(4 * 3 * 16 * 2) + 1/(16 * 4 * 9) + 1/(16 * 3 * 4 * 4)
= 1/64(1/6 + 1/9 + 1/12) = 1/64 * 13/36 of the work in a day.
They will take (64 * 36)/13 days to complete the work working together.
Train P crosses a pole in 30 seconds and train Q crosses the same pole in one minute and 15 seconds. The length of train P is threefourths the length of train Q. What is the ratio of the speed of train P to that of train Q?
Given that train P crosses a pole in 30 seconds and train Q crosses the same pole in one minute and 15 seconds.
Let the length of train P be L_{P} and that of train Q be L_{Q}
given that L_{P} = 3/4 L_{Q}
As the train P and Q crosses the pole in 30 seconds and 75 seconds respectively,
=> Speed of train P = V_{P} = L_{P}/30
Speed of train Q = V_{Q} = L_{Q}/75
L_{P} = 3/4 L_{Q}
=> V_{P} = 3/4 L_{Q}/(30) = L_{Q}/40
Ratio of their speeds = V_{P} : V_{Q}
= L_{Q}/40 : L_{Q}/75 => 1/40 : 1/75 = 15 : 8
Given that train P crosses a pole in 30 seconds and train Q crosses the same pole in one minute and 15 seconds.
Let the length of train P be L_{P} and that of train Q be L_{Q}
given that L_{P} = 3/4 L_{Q}
As the train P and Q crosses the pole in 30 seconds and 75 seconds respectively,
=> Speed of train P = V_{P} = L_{P}/30
Speed of train Q = V_{Q} = L_{Q}/75
L_{P} = 3/4 L_{Q}
=> V_{P} = 3/4 L_{Q}/(30) = L_{Q}/40
Ratio of their speeds = V_{P} : V_{Q}
= L_{Q}/40 : L_{Q}/75 => 1/40 : 1/75 = 15 : 8
The ratio between the school ages of Neelam and Shaan is 5:6 respectively. If the ratio between the onethird age of Neelam and half of Shaan’s age of 5:9, then what is the school age of Shaan?
Let the school ages of Neelam and Shaan be 5x and 6x years respectively.
Then, (1/3 * 5x)/(1/2 * 6x) = 5/9
15 = 15
Thus, Shaan’s age cannot be determined.
Let the school ages of Neelam and Shaan be 5x and 6x years respectively.
Then, (1/3 * 5x)/(1/2 * 6x) = 5/9
15 = 15
Thus, Shaan’s age cannot be determined.
At what rate percent on simple interest will Rs.750 amount to Rs.900 in 5 years?
150 = (750*5*R)/100
R = 4%
150 = (750*5*R)/100
R = 4%
A, B and C can do a work in 90, 30 and 45 days respectively. If they work together, in how many days will they complete the work?
One days’s work of A, B and C = 1/90 + 1/30 + 1/45
= (1 + 3 + 2)/90 = 1/15
A, B and C together can do the work in 15 days.
One days’s work of A, B and C = 1/90 + 1/30 + 1/45
= (1 + 3 + 2)/90 = 1/15
A, B and C together can do the work in 15 days.
– 84 * 29 + 365 = ?
Given Exp. = – 84 * (30 – 1) + 365
= – (84 * 30) + 84 + 365
= – 2520 + 449 = – 2071
Given Exp. = – 84 * (30 – 1) + 365
= – (84 * 30) + 84 + 365
= – 2520 + 449 = – 2071
A sum of money placed at C.I. interest doubles itself in 5 years. It will amount to eight times itself at the same rate of interest in?
P(1 + R/100)^{5} = 2P => (1 + R/100)^{5} = 2
Let P(1 + R/100)^{n} = 8P
=> (1 + R/100)^{n} = 8 = 2^{3} = {(1 + R/100)^{5}}^{3}
=> (1 + R/100)^{n} = (1 + R/100)^{15} => n = 15 Required time = 15 years.
P(1 + R/100)^{5} = 2P => (1 + R/100)^{5} = 2
Let P(1 + R/100)^{n} = 8P
=> (1 + R/100)^{n} = 8 = 2^{3} = {(1 + R/100)^{5}}^{3}
=> (1 + R/100)^{n} = (1 + R/100)^{15} => n = 15 Required time = 15 years.
The ratio between the radii of two spheres is 1:3. Find the ratio between their volumes?
r_{1} : r_{2} = 1:3
r_{1}^{3} : r_{2}^{3} = 1:27
r_{1} : r_{2} = 1:3
r_{1}^{3} : r_{2}^{3} = 1:27
(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
The G.C.D of 1.08, 0.36 and 0.9 is
Given numbers are 1.08, 0.36 and 0.90.
H.C.F of 108, 36 and 90 is 18.
H.C.F of a given numbers = 0.18
Given numbers are 1.08, 0.36 and 0.90.
H.C.F of 108, 36 and 90 is 18.
H.C.F of a given numbers = 0.18
Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours. C is closed A and B can fill the remaining par in 7 hours. The number of hours taken by C alone to fill the tank is?
Part filled in 2 hours = 2/6 = 1/3.
Remaining part = 1 – 1/3 = 2/3
(A + B)’s 1 hour work = 2/21
C’s 1 hour work = [(A + B + C)’s 1 hour work – (A + B)’s 1 hour work]
= (1/6 – 2/21) = 1/14
C alone can fill the tank in 14 hours.
Part filled in 2 hours = 2/6 = 1/3.
Remaining part = 1 – 1/3 = 2/3
(A + B)’s 1 hour work = 2/21
C’s 1 hour work = [(A + B + C)’s 1 hour work – (A + B)’s 1 hour work]
= (1/6 – 2/21) = 1/14
C alone can fill the tank in 14 hours.
One half of a two digit number exceeds its one third by 6. What is the sum of the digits of the number?
x/2 – x/3 = 6 => x =6
3 + 6 = 9
x/2 – x/3 = 6 => x =6
3 + 6 = 9
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?
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 = (p + q + r)/(r + q + p) = 1.
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 = (p + q + r)/(r + q + p) = 1.
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.
The list price of an article is Rs.65. A customer pays Rs.56.16 for it. He was given two successive discounts, one of them being 10%. The other discount is?
65*(90/100)*((100x)/100) = 56.16
x = 4%
65*(90/100)*((100x)/100) = 56.16
x = 4%
The difference between two integers is 5. Their product is 500. Find the numbers.
Let the integers be x and (x + 5). Then,
x(x + 5) = 500
x^{2} + 5x – 500 = 0
(x + 25)(x – 20) = 0
x = 20
So, the numbers are 20 and 25.
Let the integers be x and (x + 5). Then,
x(x + 5) = 500
x^{2} + 5x – 500 = 0
(x + 25)(x – 20) = 0
x = 20
So, the numbers are 20 and 25.
A certain sum of money at simple interest amounted Rs.840 in 10 years at 3% per annum, find the sum?
840 = P [1 + (10*3)/100]
P = 646
840 = P [1 + (10*3)/100]
P = 646
There are four prime numbers written in ascending order of magnitude. The product of the first three is 7429 and of the last three is 12673. Find the fourth number?
(7429/12673) = 17/29 Fourth Number is 29
(7429/12673) = 17/29 Fourth Number is 29
(1000)9/1024 = ?
Given Exp. = (1000)^{9}/10^{24}
= (10^{3})^{9}/10^{24} = 10^{27}/10^{24} = 10^{(27 – 24)} = 10^{3} =1000
Given Exp. = (1000)^{9}/10^{24}
= (10^{3})^{9}/10^{24} = 10^{27}/10^{24} = 10^{(27 – 24)} = 10^{3} =1000
An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m?
Length of the first carpet = (1.44)(6) = 8.64 cm
Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)
= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m
Cost of the second carpet = (45)(12.96 * 7) = 315 (13 – 0.04) = 4095 – 12.6 = Rs. 4082.40
Length of the first carpet = (1.44)(6) = 8.64 cm
Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)
= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m
Cost of the second carpet = (45)(12.96 * 7) = 315 (13 – 0.04) = 4095 – 12.6 = Rs. 4082.40
On dividing a number by 68, we get 269 as quotient and 0 as remainder. On dividing the same number by 67, what will be the remainder?
Number = 269 * 68 + 0 = 18292
67) 18292 (273
18291
——–
1
Required number = 1.
Number = 269 * 68 + 0 = 18292
67) 18292 (273
18291
——–
1
Required number = 1.
Find the size of the largest square slabs which can be paved on the floor of a room 5 meters 44 cm long and 3 meters 74 cm broad?
HCF of 374, 544 = 34
HCF of 374, 544 = 34
If a man walks to his office at ¾ of his usual rate, he reaches office 1/3 of an hour late than usual. What is his usual time to reach office?
Speed Ratio = 1:3/4 = 4:3
Time Ratio = 3:4
1 ——– 1/3
3 ——— ? → 1 hour
Speed Ratio = 1:3/4 = 4:3
Time Ratio = 3:4
1 ——– 1/3
3 ——— ? → 1 hour
On an order of 5 dozen boxes of a consumer product, a retailer receives an extra dozen free. This is equivalent to allowing him a discount of:
Clearly, the retailer gets 1 dozen out of 6 dozens free.
Equivalent discount = 1/6 * 100 = 16 2/3%.
Clearly, the retailer gets 1 dozen out of 6 dozens free.
Equivalent discount = 1/6 * 100 = 16 2/3%.
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.
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
Ajay spends 45% of his monthly income on household items, 25% of his monthly income on buying cloths, 7.5% of his monthly income on medicines and saves the remaining amount which is Rs. 9000. Find his monthly income.
Let the monthly income of Ajay be Rs. x
Savings of Ajay = x – (45 + 25 + 7.5)/100 * x = 22.5/100 x
22.5/100 x = 9000
x = 40000.
Let the monthly income of Ajay be Rs. x
Savings of Ajay = x – (45 + 25 + 7.5)/100 * x = 22.5/100 x
22.5/100 x = 9000
x = 40000.
How many of the following numbers are divisible by 132?
264, 396, 462, 792, 968, 2178, 5184, 6336
132 = 11 * 3 * 4
Clearly, 968 is not divisible by 3
None of 462 and 2178 is divisible by 4
And, 5284 is not divisible by 11
Each one of the remaining four numbers is divisible by each one of 4, 3 and 11.
So, there are 4 such numbers.
132 = 11 * 3 * 4
Clearly, 968 is not divisible by 3
None of 462 and 2178 is divisible by 4
And, 5284 is not divisible by 11
Each one of the remaining four numbers is divisible by each one of 4, 3 and 11.
So, there are 4 such numbers.
Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them?
P(at least one graduate) = 1 – P(no graduates) =
1 – ⁶C₃/¹⁰C₃ = 1 – (6 * 5 * 4)/(10 * 9 * 8) = 5/6
P(at least one graduate) = 1 – P(no graduates) =
1 – ⁶C₃/¹⁰C₃ = 1 – (6 * 5 * 4)/(10 * 9 * 8) = 5/6
(4300731) – ? = 2535618
Let 4300731 – x = 2535618
Then, x = 4300731 – 2535618 = 1765113
Let 4300731 – x = 2535618
Then, x = 4300731 – 2535618 = 1765113
P, Q and R have Rs.6000 among themselves. R has twothirds of the total amount with P and Q. Find the amount with R?
Let the amount with R be Rs.r
r = 2/3 (total amount with P and Q)
r = 2/3(6000 – r) => 3r = 12000 – 2r
=> 5r = 12000 => r = 2400.
Let the amount with R be Rs.r
r = 2/3 (total amount with P and Q)
r = 2/3(6000 – r) => 3r = 12000 – 2r
=> 5r = 12000 => r = 2400.
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};
A, B and C can do a work in 6 days, 8 days and 12 days respectively. In how many days can all three of them working together, complete the work?
Work done by all three of them in one day = 1/6 + 1/8 + 1/12 = 3/8.
The number of days required = 8/3 = 2 2/3 days.
Work done by all three of them in one day = 1/6 + 1/8 + 1/12 = 3/8.
The number of days required = 8/3 = 2 2/3 days.
The marks obtained by Vijay and Amith are in the ratio 4:5 and those obtained by Amith and Abhishek in the ratio of 3:2. The marks obtained by Vijay and Abhishek are in the ratio of?
4:5
3:2
——
12:15:10
12:10
6:5
4:5
3:2
——
12:15:10
12:10
6:5
52 * 40.1 = ?
Since, 40.1 = 40
52 * 40 = 2080
Since, 40.1 = 40
52 * 40 = 2080
The sum of money at compound interest amounts to thrice itself in 3 years. In how many years will it be 9 times itself?
100 — 300 — 3
900 — 3
—
6 years
100 — 300 — 3
900 — 3
—
6 years
A, B and C rents a pasture for Rs.870. A put in 12 horses for 8 months, B 16 horses for 9 months and 18 horses for 6 months. How much should C pay?
12*8 :16*9 = 18*6
8: 12: 9
9/29 * 870 = 270
12*8 :16*9 = 18*6
8: 12: 9
9/29 * 870 = 270
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
30.9 * 3000 – 10.1 * 1100 + 8298 – 4302 = ?
9 * 3000 – 10.1 * 1100 + 8238 – 4302
31 * 3000 – 1010 * 11 + 8298 – 4302
93000 + 8298 – 11110 – 4302 = 101298 – 15412
= 85886 = 85000
9 * 3000 – 10.1 * 1100 + 8238 – 4302
31 * 3000 – 1010 * 11 + 8298 – 4302
93000 + 8298 – 11110 – 4302 = 101298 – 15412
= 85886 = 85000
If the price of gold increases by 50%, find by how much the quantity of ornaments must be reduced, so that the expenditure may remain the same as before?
100
150
——
150——50
100——? => 331/3%
100
150
——
150——50
100——? => 331/3%
The average of 10 numbers is calculated as 15. It is discovered later on that while calculating the average, one number namely 36 was wrongly read as 26. The correct average is?