Numerical Ability
Question 1
Answer
Explanation
Question 1
For a grouped frequency distribution having eight classes, the upper class boundaries of the lowest and the highest classes are 10 and 66 respectively. What is the lower class boundary of the highest class?
A. 57
B. 60
C. 58
D. 59
Answer
C. 58
Explanation
upper class boundary of lowest class = 10
upper class boundary of highest class = 66
Total classes = 8
66 – 10 = 56
class interval = 56 / 8 + 1 = 8
lower boundary of the highest class = 66 – 8 = 58
Question 2
Answer
Explanation
Question 2
If (√7+ √5) / (√7 − √5)= a − b√35, then the value a – 2b is
A. 4
B. 5
C. 8
D. 7
Answer
C. 8
Explanation
=> (√7+ √5)^2 / 2 = a – b√35
=> (12 + 2√35) / 2 = a – b√35
=> 6 + √35 = a – b√35
a = 6, b = -1 => a-2b = 6+2=8
Question 3
Answer
Explanation
Question 3
The average marks (out of 100) scored by the students of four towns A, B, C and D of a district, in English and Hindi at a secondary level examination held in 2019 are presented through a Bar Graph shown below.
How much percentage (correct up to two decimal places) are the average marks in Hindi above that of the highest marks among the towns in English?
A. 18.33
B. 14.33
C. 16.67
D. 20.67
Answer
C. 16.67
Explanation
Avg. marks in Hindi = (60+80+70+70)/4 = 70
Highest marks in English = 60
%above = (70-60)*100/60 = 16.67
Question 4
Answer
Explanation
Question 4
Raju buys 3 goats and 2 sheeps for Rs.11600. When he sells the goats at 20% profit and the sheep at 10% loss, he earns a total profit of Rs.1000. The cost of one sheep is _____.
A. Rs.2600
B. Rs.4600
C. Rs.2400
D. Rs.2200
Answer
D. Rs.2200
Explanation
Let, CP of one goat is Rs. x and CP of one sheep is Rs. y
3x + 2y = Rs.11600 …(1)
Now, given that, he sells goat for 20% profit and sheep at 10% loss.
SP of 3 goats = (3x * 120/100) = Rs.3.6x
SP of 2 sheep = (2y * 90)/100) = Rs.1.8y
Total Profit is = Rs.1000
Total SP = 11600 + 1000 = Rs.12600
3.6x + 1.8x = 12600 …(2)
Solving Eqn.(1) and Eqn.(2) we get,
=> 10(3.6x + 1.8x) – 12(3x + 2y) = 10*12600 – 12*11600
=> 36x – 36x + 18y – 24y = 126000 – 139200
=> (-6y) = (-13200)
=> y = Rs.2200
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Question 5
Answer
Explanation
Question 5
The sum of two numbers is 2604 and their HCF is 124. Which is the smaller between them if their difference is the least possible?
A. 1116
B. 1240
C. 620
D. 496
Answer
B. 1240
Explanation
x + y = 2604
HCF is 124
For difference to be minimum y=x+124 (Assuming y to be larger number)
Then we get
=> x+x+124=2604
=> 2x=2480
=> x=1240 (minimum number is x)
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Question 6
Answer
Explanation
Question 6
What sum (in Rs) given on loan for two years with scheme of return on the basis of compound interest at a yearly rate of 10% will correspond to repayment through equal monthly installments of Rs.9075?
A. 165000
B. 180000
C. 189000
D. 198000
Answer
B. 180000
Explanation
Total repayment = Rs. 9075 * 24 = Rs. 2,17,800
Let, loan amount = Rs. P
ATQ, P*(1+10/100)^2 = 217800
=> P*(121/100) = 217800
=> P = 180000
Question 7
Answer
Explanation
Question 7
The capacities of three containers X, Y and Z are 1, 2 and 4 liters respectively. Initially, X is empty, while Y and Z are full of water and milk respectively. X is filled from Y, Y is replenished from Z and X is emptied into Z. If this process is repeated once more, then what will be the ratio
of milk in Y to water in Z?
A. 3:5
B. 1:1
C. 4:3
D. 4:5
Answer
B. 1:1
Explanation
X(1) Y(2) Z(4)
Initially M=0 W= 0 M=0 W = 2 M = 4 W = 0
X is filled from Y M=0 W= 1 M=0 W = 1 M = 4 W = 0
Y is filled from Z M=0 W= 1 M=1 W = 1 M = 3 W = 0
X is emptied in Z M=0 W= 0 M=1 W = 1 M = 3 W = 1
Milk and water in Y are in 1:1 ratio hence 1 litre will have 0.5 Water & 0.5 milk
X is filled from Y M=0.5 W= 0.5 M=0.5 W = 0.5 M = 3 W = 1
Milk and water in Z are in 3:1 ratio hence 1 litre will have 0.25 Water & 0.75 milk
Y is filled from Z M=0.5 W= 0.5 M=1.25 W = 0.75 M = 2.25 W = 0.75
X is emptied in Z M=0 W= 0 M=1.25 W = 0.75 M = 2.75 W = 1.25
Milk in Y = 1.25
Water in Z = 1.25
1.25 : 1.25 = 1 : 1
ratio of milk in Y to water in Z = 1 : 1
Question 8
Answer
Explanation
Question 8
The average score in Mathematics of a class increases by 10% if the total marks secured by a number of students who form 20% of the class strength and whose average score is 48 is not included in the calculation. What is the average score?
A. 90
B. 60
C. 75
D. 80
Answer
D. 80
Explanation
Let average score = x, Total students = 100
Total marks = 100x
0% of Total students = 20 students
Their average score = 48 marks
Total marks not included = 48 * 20 = 960 marks
Marks left now = (100x – 960) marks
Students left = 100 – 20 = 80 students
New average score = {(100x – 960)/80} marks
Given that, average score is increased by 10%
(100x – 960) / 80 = (110x/100)
=> 5(100x – 960) = 4 * 110x
=> 500x – 5 * 960 = 440x
=> 500x – 440x = 5 * 960
=> 60x = 5 * 960
=> x = 80
Question 9
Answer
Explanation
Question 9
If 25% of a number is equal to the three-fifths of another number, what will be the ratio of the first number to the second number?
A. 5:12
B. 12:5
C. 12:15
D. 15:12
Answer
B. 12:5
Explanation
First Number = x, Second Number = y
ATQ, x/4 = 3y/5
=> x/y = 12/5 = 12:5
Question 10
Answer
Explanation
Question 10
A hollow spherical ball of thickness 1 cm and external radius 5 cm is melted and then from the solid so obtained, without any loss of material, 61 identical spherical balls are obtained. What is the diameter (in cm) of each ball?
A. 1.5
B. 3
C. 1
D. 2
Answer
D. 2
Explanation
Internal radius = 5-1 = 4cm
Volume of the melted material = (4/3)*pi*(5^3-4^3) = (4/3)*pi*61 cc
61 indentical ball are obtained from this.
Volume of each ball = (4/3)*pi cc
So, radius of each ball = 1cm
Diameter of each ball = 2cm
Question 11
Answer
Explanation
Question 11
Four distributions AddaJobs, Peepinsta, PhraseSeven, Offlineprepare9i are given below:
AddaJobs: 4, 2, 1, 3, 3, 5, 6, 9, 8, 7
Peepinsta: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
PhraseSeven: 5, 6, 4, 5, 7, 5, 8, 1, 5, 2
Offlineprepare9i: 2, 6, 6, 11, 10, 9, 10, 8, 7
What is the mean of the two modes in the only bimodal distribution among the above?
A. 5
B. 3
C. 6
D. 8
Answer
D. 8
Explanation
Offlineprepare9i is the only bimodal distribution here.
It has 6 and 10 two times, which makes them the mode.
The mean of the mode = (6+10)/2 = 8
Also Checkout: TCS NQT Sample Question Paper
Question 12
Answer
Explanation
Question 12
What is the value of 1.59 X 1.59 + 8.46 X 0.53 + 9 X 0.47 X 0.47 ?
A. 9.025
B. 6.25
C. 9
D. 4
Answer
C. 9
Explanation
1.59 * 1.59 + 8.46 * 0.53 + 9 * 0.47 * 0.47
= 9 * 0.53 * 0.53 + 9 * 0.94 * 0.53 + 9 * 0.47 * 0.47
= 9 * (0.53^2 + 2 * 0.47 * 0.53 + 0.47^2)
= 9 * (0.53+0.47)^2
= 9 * 1 = 9
Question 13
Answer
Explanation
Question 13
What is the simplified value of: (7^2 + 24^2) / [20^2 – (1/2)(5^2 – 1) + 10{(0.3)^2 + (0.1)^2} – 2 x 7]^2?
A. 1/5
B. 1/225
C. 1/125
D. 1/25
Answer
B. 1/225
Explanation
(7^2 + 24^2) / [20^2 – (1/2)(5^2 – 1) + 10{(0.3)^2 + (0.1)^2} – 2 x 7]^2
= (49 + 576) / [400 – (1/2)24 + 10{0.09 + 0.01} – 14]^2
= (625) / [400 – 12 + 1 – 14]^2
= (625) / (375 * 375) = 1/225
Question 14
Answer
Explanation
Question 14
Four men and two women can do a piece of work together in one day. If a woman is twice as efficient as a man, in how many days can a woman working alone do the work?
A. 4
B. 6
C. 8
D. 2
Answer
A. 4
Explanation
Let, Efficiency of a man is x unit/day.
Efficiency of a woman = 2x unit/day.
4 men completes = 4x unit of work / day.
2 women completes = 4x unit of work / day.
1 one day total unit of work completes by 4 men and 2 women is = 4x + 4x = 8x units. = Total work
Time taken by a woman working alone to do the complete work = 8x/2x = 4 days.
Question 15
Answer
Explanation
Question 15
The median of the following data is ______.
A. 24.266
B. 24.266
C. 24.123
D. 24.166
Answer
D. 24.166
Explanation
C.I. (m) (f) (d) fd c.f.
0-10 5 10 -2 -20 10
10-20 15 15 -1 -15 25
20-30 25 12 0 0 37
30-40 35 15 1 15 52
40-50 45 8 2 16 60
Question 16
Answer
Explanation
Question 16
A train starting from station X was to arrive at station Y at 6:06 PM. It could travel at 62.5% of its usual speed and reach Y at 7 PM. At what time did it start from X?
A. 4:44 PM
B. 4:56 PM
C. 4:36 PM
D. 4:24 PM
Answer
C. 4:36 PM
Explanation
Let, the actual speed of the train be 100x km/hr and the actual time taken be y hours.
Distance covered = (100xy)km
now, given that, if train travels at 62.5% of its usual
New speed = 62.5% of 100x = 62.5x km/h
increased time = 7PM – 6:06 = 54 minutes = (54/60) = (9/10) hours
New total time = (y + 9/10) hours
Distance covered = (62.5x) * {y + (9/10)} km
=> 100xy = 62.5x * {y + (9/10)}
=> 100y = 62.5(10y + 9)/10
=> 1000y = 625y + 562.5
=> 1000y – 625y = 562.5
=> 375y = 562.5
=> y = 1.5 hours.
Train start from X at = 6:06 – 1:30 = 5:66 – 1:30 = 4:36 PM.
Question 17
Answer
Explanation
Question 17
In how many ways can X give Rs.500 to Y using only Rs.100 and Rs.20 notes, with the condition that she has only 16 notes of Rs.20, and being asked to use the notes of both the denominations?
A. 6
B. 3
C. 4
D. 2
Answer
B. 3
Explanation
5 Rs.20 notes + 4 Rs. 100 notes = 500
10 Rs.20 notes + 3 Rs. 100 notes = 500
15 Rs.20 notes + 2 Rs. 100 notes = 500
therefore, 3 ways.
Question 18
Answer
Explanation
Question 18
A sales representative’s commission is 6% on all sales up to Rs. 15000 and 5% on all sales exceeding this. He remits Rs. 47350 to his company after deducting his commission. What were the total sales?
A. Rs. 49000
B. Rs. 47500
C. Rs. 50500
D. Rs. 50000
Answer
D. Rs. 50000
Explanation
Let, Total Sales = x
Total Commission = y
Then according to the given conditions,
0.06*15000 + 0.05 (x – 15000) = y …1
47350 + y = x …2
On simplifying 1, we have
900 + 0.05x – 750 = y
Therefore, y – 0.05x = 150 ………E3
On simplifying E2, we have
x – y = 47350 ………E4
To solve this, adding E3 & E4, we get
0.95x = 47500
Therefore, x = 50000
Hence, the total sales is Rs. 50000
Question 19
Answer
Explanation
Question 19
A particular distribution is represented by two data points. If the range and the standard deviation of the distribution are R & S respectively, what is the relation between them?
A. S = √R
B. S = 2R
C. S = R
D. S = R/2
Answer
D. S = R/2
Explanation
Let, the points are a,b.
Range, R = a-b
SD, S = √[(1/2){(a-(a+b)/2)^2 + (b-(a+b)/2)^2}]
= √[(1/2){(a/2-b/2)^2 + (b/2-a/2)^2}]
= √[(1/2){(a/2-b/2)^2 + (a/2-b/2)^2}]
= √[{(a-b)/2}^2] = (a-b)/2
Therefore, S = R/2
Also Checkout: TCS NQT Sample Question Paper
Question 20
Answer
Explanation
Question 20
What will be the percentage increase in the area of a square, If its side is increased by 20%?
A. 44%
B. 40%
C. 36%
D. 20%
Answer
A. 44%
Explanation
Let, initial length of side be x
Increased length = x+20%(x) = 1.2x
Initial area =x^2
Increased area = (1.2x)^2 = 1.44x^2
Percent increase in area= (1.44x^2-x^2)*100/x^2 =0.44×100=44%
Question 21
Answer
Explanation
Question 21
The ratio is incomes of P and Q is 7:5 and the ratio of their expenditures is 4:3. IF at the end of the year, P and Q save Rs.3000 and Rs. 2000 respectively, what is Q’s income?
A. Rs.5000
B. Rs.4500
C. Rs.4000
D. Rs.7000
Answer
A. Rs.5000
Explanation
Ratios of income = 7:5 => 7x and 5x
Ratios of exp = 4:3 => 4y and 3y
Savings of P = 3000
Savings of Q = 2000
So, 7x – 4y = 3000
5x – 3y = 2000
On solving we get x = 1000 and y = 1000
Q’s income = 5 * 1000 = 5000
Question 22
Answer
Explanation
Question 22
Raju lends Rs. 3000 to Bharath and a certain sum to Charan at the same time at 6% per annum simple interest. If after 5 years, Raju altogether receives Rs. 1650 as the interest from Bharath and Charan, what is the sum lent to Charan?
A. Rs. 2500
B. Rs. 2750
C. Rs. 3250
D. Rs. 3300
Answer
A. Rs. 2500
Explanation
Case 1> Sum Lends to Bharath.
P = Rs.3000
R = 6% per annum.
T = 5 years.
SI = (3000 * 6 * 5) / 100 = Rs.900
So, Raju recieved Rs.900 as interest from Bharath.
Case 2> Let sum lends to Charan is Rs.x .
P = Rs. x
R = 6% per annum.
T = 5 years.
SI = (x * 6 * 5) / 100 = Rs.(3x/10)
So, Raju recieved Rs.(3x/10) as interest from Charan.
Therefore,
900 + (3x/10) = 1650
=> (3x/10) = 1650 – 900
=> 3x = 7500
=> x = Rs.2500
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