# TCS NQT Numerical Ability Questions & Answers Set 3 : TCS NQT Placement Papers 6

Here the latest TCS NQT Quantitative, Numerical Ability Questions and Answers with solutions. You can download this TCS NQT Prevuiuos year tcs nqt question paper in pdf by clicking on Print option on your browser.

These are the official TCS Off campus drive question paper of Numerical Ability exam held in October 2020.

## NQT Numerical Ability Question Paper Set 3:

### Question 1: In case of frequency distribution of ten continuous classes, the class width is 4 and the lower class limit of the lowest class is 8. What is the upper class limit for the highest class?

1. 48
2. 38
3. 46
4. 34

Solution : Option 1

Let x and y be the upper and lower class limit of frequency distribution.
Given, width of the class =4
− y=4……….(I)
Also, given lower class (y)=8
On putting y =8 in Eq. (i), we get
− 8 = 4 ⇒ = 12
So, the upper class limit of the lowest class is 12.
Hence, the upper limit of the highest class= (Number of continuous classes x Class with + Lower class limit of the lowest class)
= 10 × 4 + 8 = 40 + 8 = 48
Hence, the upper class limit of the highest class is 48.

### Question 2: What is the simplified value of (1.728+1/1.24) + (1-0.343/2.19)

A. 2.5
B. 1.5
C. 3.5
D. 4.5

Solution: Option A

(1.728+1/1.24) + (1-0.343/2.19)

2.728/1.24 + 0.657/2.19

2.2 + 0.3 = 2.5

### Question 3: A leak in the body of a ship that can admit 5 units of water in 16 minutes got detected when it was 40 km away from the shore. The entry of 60 units of water will cause the ship to sink which has a pump that can throw out 12 units of water in an hour. Aided by the pump, what should be the average speed (im Kmph) of the ship in order to reach the shore just before it starts sinking?

1. 4
2. 4.2
3. 4.5
4. 4.8

Solution: Option 4

given that,
→ in 16 minutes water is entering inside ship = 5 units .
so,
→ in 1 minute water is entering inside ship = (5/16) units.
also,
→ in 60 minutes water thrown out of the ship = 12 units.
→ in 1 minute water thrown out of the ship = (12/60) = (1/5) units.
then,
→ in 1 minute water entering inside ship = (5/16) – (1/5) = (25 – 16)/80 = (9/80)units.
Therefore,
→ Time taken by ship before sink = (maximum water ship can hold before sink) / (water entering inside ship per minute) = 60/(9/80) = (60 * 80)/9 = (1600/3) minutes = (1600/3) * (1/60) = (80/9) hours.
Hence,
Average speed of ship = (Distance from the shore /Time taken by ship before sink) = 40 / (80/9) = (40 * 9)/80 = 4.5km/h.
the average speed (im Kmph) of the ship in order to reach the shore just before it starts sinking=4.5km/h

### Question 4 :What is the fourth proportional of the mean proportionals between (4, 6.25): (0.0027, 300) :: (1..2 x 10-3, 3) : ?

A.0.0108
B.0.00108
C.0.000108
D. 0.1008

Solution: Option A

(4, 6.25) = a

(0.0027, 300) = b

(1..2 x 10-3, 3) = c

? = d

a: b :: c: d

Mean Proportional of a =√4 X 6.25 = 5

Mean Proportional of b=√0.0027 X 300 = 0.9

Mean Proportional of c =√1.2 X 10^-3 X 3 = 0.06

a/b = c/d

5/0.9 = 0.06/d

d= 0.06 X 0.9/5

d= 0.0108

### Question 5 :Instead of calculating the sum of a proper fraction, less than 1/2 , with its reciprocal, the difference was worked out, as a result of which there was an error of 121(39/41)%. What was the fraction?

1. 5/3
2. 3/5
3. 5/4
4. 4/5

Solution: Option 4

Let us assume that, the given proper fraction is (x/y) where y > x.

So,

→ Actual sum = (x/y) + (y/x). = (x² + y²)/xy

but,

→ calculated value = (x/y) – (y/x) = (x² – y²)/xy

So,

→ Error = Actual sum – calculated value = {(x² + y²)/xy} – {(x² – y²)/xy} = {(x² + y² – x² + y²) / xy} = 2y²/xy = (2y/x).

than,

→ % error = {2y/x) * 100 / [ (x² + y²)/xy ] = (200y * xy) / x(x² + y²) = (200y²/(x² + y²) %

given that, % error was 121(39/41) % .

therefore,

→ 200y²/(x² + y²) = 121(39/41)

→ 200y²/(x² + y²) = 5000/41

→ y²/(x² + y²) = 25/41

→ y²/(x² + y²) = 25 / (25 + 16)

→ y²/(x² + y²) = 5² / (5² + 4²)

comparing , we get,

→ y = 5.

→ x = 4.

Hence, Required Proper Fraction is = (4/5)

### Question 6: A retailer purchases several pens of the same cost for Rs 384 and then sells some of them for Rs 114 at no loss or gain. What is the least possible number of pens that he will be left with?

A. 46
B. 45
C. 44
D. 47

Solution : Option B

Prime factorisation of 384 and 114 gives:

384 = (2^7) * 3

114 = 2 * 3 * 19

HCF (384, 114) = 2 * 3 = 6.

Thus, the highest amount that every pen could cost = 6 rupees

The retailer purchased (384 / 6) pens = 64 pens and sold (114 / 6) pens = 19 pens from those.

Therefore, least number of pens left with the retailer now = (64 – 19) = 45.

### Question 7: A sales representative commission is 6% on all sales up to Rs 15,000 and 5% on all the sales exceeding this. He remits Rs 47,350 to his company after deducting  his commission. What was the total sale?

1. 49000
2. 50500
3. 50000
4. 47500

Solution: Option C

47350 = x – 6% of 15000 – 5% of (x-15000)

47350 = x – 900 – 5% of x – 5% of 15000

47350 + 900 – 750 = x – 5% of x

47500 = 95% of x

=> x = 47500 x 100/95

x = 50000

Rs. 50000

### Question 8: The area of the sector with central angle 150, of a circle is 231 sq cm. If pi = 22/7, what is the circumference (in cm) of the circle?

A. 264
B. 254
C. 244
D. 285

Solution:

Area of Sector = 𝞹 r2 (Sector Angle/360) = 231

22/7 * r2 * (15/360) = 231

22/7 * r2 * (1/24) = 231

r2 = ((231 * 24 * 7)/22)

r2 = 21 * 12 * 7

r2 = 1764

r = 42

Circumference of circle = 2𝞹r

2 * 22/7 * 42

= 22 * 12

= 264 cm

### Question 9: In a competitive exam, 5 marks are awarded for every correct answer and for every wrong answer, 2 marks are deducted. Sathwik scores 32 marks in the examination. If 4 marks had been awarded for each correct answer and 1 mark had been deducted for each incorrect answer, Sathwik would have scored 34 marks. If Sathwik attempted all the questions, how many questions were there in the test ?

1. 14
2. 12
3. 20
4. 26

Solution: Option 4

In  given Question:-
So,
5x -2y = 32
4x-2y =34——*2
Now
5x-2y =32
8x-4y=68
By solving
3x=36
x=12
Now to find ‘y’
5*12 -2y =32
60 -2y= 32
y=14
12+24 =26
Questions in the test =26

### Question 10: A takes 10 days less than the time taken by B to finish a piece of work If both A and B together can finish the work in 12 days the time taken by B to finish the work is

A. 20 Days
B. 30 Days
C. 40 Days
D. 50 Days

Solution: Option B

Let B do the work independently in x no.s of days then

According to the question

A will take ( x – 10 ) no.s of day

Then in a unit day

A works 1/part of work and B works 1/(x-10) part of work, as they both complete it in 12 days then

1/ x + 1/(x -10) =1/12 (portion of work in a day)

x(x-10)/2x -10 = 12

x^2 – 10x = 24x + 120

(x-30) (x-4)

Possible value is x=30

Hence B will do the work in 30  days

### Question 11:  What is the simplified value of

1. 6
2. 4
3. 2
4. 5

Solution: Option 4

=5

### Question 12: The total surface area of a solid hemisphere is 75 times the area of a circle whose diameter is 6 cm. What is the radius (in cm) of the hemisphere?

1. 10 cm
2. 15 cm
3. 20 cm
4. 25 cm

Solution : Option 2

area of circle = πr2 = = π x 32=9π

Total surface area of hemisphere = 75 x 9π

Now , 3πr2 = 75 x 9π

= r2 = 75 x 3= 225

r = 15 cm

### Question 13: 10 men can complete a work in 8 days and 10 women take 12 days to complete the same work. How many days will 3 men and 6 women take to complete the same work?

A. 9 3/7
B. 10 3/7
C. 11 3/7
D. 12 3/7

Solution : Option C

1 men 1 day’s work = 1/80
1 men 1 day’s work = 1/120
(3 men + 6 women )’s day’s work = ( 1/80 + 1/120) = 7/80

5 men and 10 women will complete the work in 80/7 days. = 11 3/7 days

### Question14 :The ratio of the average speeds of two sprinters is 8:7. In a racing track if  B starts at a distance of 2m ahead of A. At what distance (in metre) from the starting point of B would A catch up with B?

A.14 m
B. 15 m
C.16 m
D.17 m

Solution :Option C

As we know that,

• A gives a start of x metres to B implies if A starts the race from starting point, then B starts x metres ahead of A. To cover race of 100 metres in this case, A will have to cover 100 metres while B will have to cover only (100 – x) metres.
• They meet for the first time on the circular track (not necessarily at the starting point) = Length of the track / Relative Speed .

So, Let us assume that, Speed of A is 8x m / sec. and speed of B is 7x m / sec.

Now, we have given that, B starts at a distance of 2m ahead of A .

we can say that, A has have to cover this 2m distance in order to catch B .

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So,

→ Distance = 2m .

→ Usual speed = (8x – 7x) = x m/sec. { same direction. }

Than,

→ Time taken by A to catch/meet B = Distance covered / usual speed = (2/x) seconds.

Therefore,

→ in (2/x) seconds A covered = Speed * Time = 8x * (2/x) = 16m.

Hence,

→ Distance (in metre) from the starting point of B would A catch up with B is = 16 – 2 = 14 m

### Question 15: Raju lends Rs 3000 to Bharath and a certain sum to Charan at the same time of 6% per annum simple interest.If after 5 years ,Raju altogether receives Rs 1650 as interest from Bharath and Charan, What is the sum lent to charan ?

1. Rs 2500
2. Rs 3300
3. Rs 2750
4. Rs 3250

Solution : Option 1

For Bharath and Charan 6% for 5 year = 30% for each
30% for 3000 = 900
=> total interest – 900 = Charan
1650 – 900 = 30% of x (given to Charan)
30% of x = 750
X = Rs 2500

### Question 16: What is the difference between the Range and the Standard Deviation of all natural numbers between 81 and 90 , both inclusive?

A.7
B.11
C. 8
D.9

Solution: Option A

range of numbers 81,82,83,84,85,86,87,88,89,90,91 is (maximum – minimum) (i.e.,) 91-81 = 10
and the standard deviation is 3.3 (approx) it is 3
difference is 10-3=7

### Question 17: The value of 1 gm gold is 90 times that of 1 gm silver. If the weights of equal volumes of silver and gold are in the ratio 10:19, what is the value of  a bar of silver that is equal in bulk to a bar of gold of value Rs 99180.

A. 560
B. 570
C. 580
D. 590

Solution: Option C

given that,

• value of 1gm gold : value of 1gm silver = 90 : 1 .
• weight of silver : weight of gold = 10 : 19.
• value of a bar of bold = Rs.99180 .

• Let us assume that,
• value of 1 gm gold = Rs.90x
• value of 1gm of silver = Rs.x.
• weight of gold = 19ygm.
• weight of silver = 10ygm.
Now,
→ Value of 1 gm of gold = Rs.90x

→ value of 19ygm of gold = Rs.(19y * 90x)

So,

Rs.(19y * 90x) = Rs.99180

→ xy = (99180) / (19 * 90)

→ xy = 58.

Therefore,

cost of 1 gm of silver = Rs.x

→ cost of 10y gm of silver = Rs.(10xy) = 10 * 58 = Rs.580

Hence the value (in Rs) of a bar of silver is 580

### Question 18:When the list price (L.P) of an article is fixed at 30% above the cost price (C.P), and while selling the same it is subject to a discount of x%, the profit percentage is 80% of that when the L.P is fixed at 25% above the C.P and the discount is 16%. What is the value of x?

A. 20%
B. 30%
C. 15%
D. 25%

Solution: Option A

Farmula : Selling price = cost price + profit
Selling price = List price – Discount
We calculate profit % on cost price and discount% on list price.

Case 1 : list price 30% more than cost price; discount x%

Case 2 : list price 25% higher than cost price and discount is 16 %

Let us assume

CP = 100 ₹

List Price 1 = 130 ₹ (30% higher than cost price)

List Price 2 = 125 ₹ (25 % higher than cost price)

Let us take second case

16 % discount

Selling price 2 = 125 * (100–16) /100

Selling price 2= 125 * 84/100 = 105

Now profit 2 = 105–100

profit 2 = 5 ₹

Profit 1 is 80% of profit 2

Profit 1 = 5 * 80/100 = 4 ₹

Selling price 1 = 100 + 4 = 104 ₹

Discount 1 = 130 -104 = 26 ₹

Discount 1 % = (discount 1/list price 1 ) *100

Discount 1 % = (26/130) *100 = 20 %

Which is equal to x % = 20 %

### Question 19:A shopkeeper calculated his profit as 12% with the selling price of an article as the base. What would have been his actual profit percentage if the selling price was 20% more?

A. 34
B. 24
C.14
D. 44

Solution: Option A

let cost price = 100

20 % more of cost price is selling price i.e 100+20% = 120

profit is calculated by taking selling price as base so – 120 + 12 % = 134.4

Therefore profit over cost price = 34.4

### Question 20: The sum of the ages of two friends is 74. After one year the ratio of their ages will be 9:10. What was the ratio of their ages nine years ago?

A.13:15
B. 13:14
C. 14:13
D. 15:13

Solution: Option A

x + y = 74

X +1 /Y+1 =9/10

10x + 10 =9y + 9

10x = 9y + 9 -10

10x = 9y-1

X = 9y -1 /10

Now putting value of x in equation 1

(9y -1 /10) +y -74

9y -1 +10y = 740

19y =741

y= 741 /19 =39

And x = 74 -y

= 74 – 39 =35

Ratio of ages nine years ago =  35-9 /39-9

=26/30

=13:15