TCS NQT Numerical Ability 2022

Let the C.P of each copy be Rs 100
ATQ one copy is sold at 0.8 x 100 = 80
One copy is sold at 1.4 x 100 = 140
Total money spent in buying the copies = Rs 200
Total money earned after selling the copies = Rs (80 + 140) = Rs 220
Total gain of Rs 20. Therefore gain% = 10%.

Area of the sector when the arc is given = (L x r)/2
Where L = length of arc and r = radius of the circle
Here L = 8 cm and r = 15 cm
Thus, Area = \frac{8 \times 15}{2} = 60 sq. cm

Prime factors of 180 and 432 is :-
→ 180 = 2 * 2 * 3 * 3 * 5
→ 432 = 2 * 2 * 2 * 2 * 3 * 3 * 3
HCF = 2 * 2 * 3 * 3 = 36.
Now , given that, HCF is written as 180a+432b.
So,
→ 180a + 432b = 36 .
→ 36(5a + 12b) = 36
dividing both sides we get,
→ 5a + 12b = 1
Now, we have given that, m and n both are integers.
So,
putting values of m , we get :-
a = 0, b = (1/12) => b = not an integer.
a = 1 , b = (-1/3) => b = not an integer.
a = 2 , b = (-3/4) => b = not an integer.
a = 3 , b = (-7/6) => b = not an integer.
a = 4 , b = (-19/12) => b = not an integer.
a = 5 , b = (-2) => b = an integer.
Hence,
→ Difference between a and b = a - b = 5 - (-2) = 5 + 2 = 7

For Amitabh and Ranbeer 6% for 5 year = 30% for each
30% for 3000 = 900
=> Total interest - 900 = Ranbeer
1650 - 900 = 30% of x (given to Ranbeer)
30% of x = 750
x = Rs 2500

Let the total number of residents be x.
As 10% of the residents did not cast their votes
Then, Votes polled =90% of x.
20% of the votes polled were found invalid
Valid votes =80% of (90% of x).
60% of [80% of (90% of x)]−40% of [80% of (90% of x)] = 2880
=>8% of [90% of (90% of x)]=1620
Solving this equation,
x= 20000

Aman’s work = 1/12 th of total work
Bibhash’s work = 1/16th of total work
Combined work of Aman and Bibhash in 1 day = 1/12 + 1/16 = 7/48
Given that they together finished their work in 6 days = ½ + 1/16 - 1/6 = -1/48
Thus, Chandan can finish the work in 48 days.
Chandan’s share = \frac{1}{48} \times 6 \times 720 = Rs 90

X can do 25% of a work in = 4 days.
So,
→ X can do 1% of a work in = (4/25) days.
Then,
→ X can do 100%(total work) of a work in = (4/25) * 100 = 16 days.
Similarly, given that,
→ Y can do (1/3) of a work in = 4 days.
Then,
→ Y can do whole work in = 4 * (3/1) = 12 days.
now, given that,
Z can complete the work in the same time as X andY would take working together.
So,
→ Time taken by Z to complete the whole work = 1 / {(1/16 + (1/12)} days.
Now, we have to find how many days all three of them will complete the same work.
So,
→ per day work of X + per day work of Y + per day work of Z = Per day work of all three.
Putting all values we get:-
→ (1/16) + (1/12) + {(1/16) + (1/12)}
→ (1/16 + 1/16) + (1/12 + 1/12)
→ (2/16) + (2/12)
→ (1/8) + (1/6)
→ (3 + 4)/24
→ (7/24)
Hence,
→ Time taken by all three to complete total work = (24/7) days.
∴ If all the three work together, they will take (24/7) days to complete the same work.

Let total number of weight of 18 students = x
Average weight = x/18=45
=> x=810
Therefore total weight of 18 students is 810kg
Now,
Adding 64 we get 874 kg total weight
So, the average weight of 19 students is 874/19=46kg

Relative speed of ants = 3 + 5 mmps = 8mmps
Effective distance = 1.1 + 1.3 = 2.4 cm = 22 mm
Time = distance/speed = 24/8 = 3

Let’s take X & Y as their ages.
Sum of their ages
x + y = 74
After 1 year, their ratio is 9/10
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

Let us assume that N alone can complete the project alone in x days.
Then, M will complete the work alone in (x + 75) days.
Given that, both take 12 days to complete the project .
So,
→ 1/x + 1/(x + 75) = 1/20
→ (x + 75 + x)/x(x + 75) = 1/20

→ (2x + 75) * 20 = x² + 75x
→ 40x + 1500 = x² + 75x
→ x² + 75x - 40x - 1500 = 0
→ x² + 35x - 1500 = 0
→ x² + 60x - 25x - 1500 = 0
→ x(x + 60) - 25(x + 60) = 0
→ (x + 60)(x - 25) = 0
Putting both equal to zero , we get,
→ x = (-60) or 25.
since days cant be negative.
Therefore,
→ Time taken by Y alone = 25 days.
→ Time taken by X alone = 25 + 75 = 100 days.
Now, we know that,
Efficiency is indirectly proportional to time..
Wages are distributed among workers in ratio of their efficiency of work per day.
So,
→ Time taken by M : Time taken by N = 100 : 25 = 4 : 1 .
Then,
→ Efficiency of M : Efficiency of N = 1 : 4.
Hence,
→ Wages of M : Wages of N = 1 : 4

Alloy M :-

Lead : copper = 3 : 7
Lead = 80 * (3/10) = 24 kg .
Copper = 80 * (7/10) = 56 kg.
Alloy N :-

Lead : copper = 7 : 3
Lead = 50 * (7/10) = 35 kg .
Copper = 50 * (3/10) = 15 kg.
Therefore,

→ Lead in new alloy = Lead in alloy M + Lead in alloy N = 24 + 35 = 59 kg.

Let the cost price be 100
Then marked price = 100 + ( 100* 40/100)
= 140
Discount percent = 25%
Price after discount = 105
Profit percentage = 5/100 * 100
= 5% Profit

Assume the cost price was 100
Then after 12% profit the selling price would have been =100+12 = 112
Now,
Profit at 20% would be= 112 x 20 /100 =22.5

Let r is the rate of interest per annum and p be the principal
157304=P(1+r/400)^{7} (A)
140000=P(1+r/400)^{5} (B)
Dividing (A) by(B)
\frac{157304}{14000}=(1+r/400)^{2}
1.1236=(1+r/400)^{2}
(1+r/400)=✓1.1236
1+r/400=1.06
r/400=1.06–1=0.06
r=24% per annum

Let us assume that, Principal is Rs. x and Let rate of interest is R% annually .
Than from given data we have :-
Principal = Rs.x
Rate = R% per annum.
Time = 3 years.
SI = 45% of x .
we know that ,
SI = (P * R * T) / 100
Putting all values we get,
→ 45% of x = (x * R * 3)/100
→ (45 * x)/100 = (x * R * 3)/100
→ 45 = 3R
dividing both sides by 3 ,
→ R = 15% .

Now, we have to find compound interest of Rs.12,000 at the same rate as 15% after 2 years.
we know that,
CI = P[{1 + (R/100)}^{time} - 1]
Putting values now we get,

→ CI = 12000[{ 1 + (15/100)}² - 1]
→ CI = 12000[{1 + (3/20)}² - 1]
→ CI = 12000[(23/20)² - 1]
→ CI = 12000[(529/400) - 1]
→ CI = 12000[(529 - 400)/400]
→ CI = 12000 * (129/400)
→ CI = 30 * 129
→ CI = Rs.3,870

Distance = Speed x Time
Speed of cycle = A km//hr
Speed of Bus = 3A km//hr
Distance = 5 km
Time taken by Cycle= 5/A hr
Time taken by Bus = 5/3A hr
5/A - 5/3A = 20/60
=> 5/A - 5/3A = 1/3
=> 15 - 5 = A
=> 10 = A

The average speed of the cycle is 10 km/hr

Total time Aashay has = 2 hrs
Total distance he has to covered = 50 km
in first 1 hr he completed 20 km
So S1= 20 km/hr
So remaining distance = 50-20 = 30 Km
And remaining time = 1 hr
Now he spent 10 minutes for refueling
So total remaining time he has = 60-10 = 50 minutes
He has to complete 30 km in this time
Speed = 30/(10/60) km/hr
Speed = 36 km/hr
For making a factor of initial speed as 20 km/hr
Multiply 36 with 20/20
Speed = 36x(20/20)
=(1.8x S1 )km/hr
= 1.8 km/hr

Let X & Y denote the amounts (in liter) of two varieties of fruit juice priced at Rs. 240 and Rs. 280 per kg respectively should be mixed so that the cost of the combination is Rs. 254 per kg.
Hence from above data we get following relation,
240*X + 280*Y = (X + Y)*254
or (254 - 240)*X = (280 - 254)*Y
or 14*X = 26*Y
or X/Y = 26/14 = 13/7
or X : Y = 13 : 7
Therefore it is evident from above that the required ratio of two varieties of fruit juice is priced at Rs. 240 and Rs. 280 per kg respectively should be mixed so that
the cost of the combination is Rs. 254 per litre = 13: 7

Let the original cost of the sketch pen set for X be “x”

X sold the item at 15% profit.
The cost for K is 1.15x, which is sold at a profit of 20%
The cost for M is 1.2(1.15x)
M paid 690
So, 690 = 1.2 \times 1.15x
Solving the equation ----
x= 500