# Igate Aptitude Placement Papers - Igate Aptitude Interview Questions and Answers updated on 16.Aug.2022

Let the number to be added be x

Then

7+x13+x=23

(7+x)3=2(13+x)

3x-2x=26-21

x=5

Hence the required number is 5

Let the required numbers be 3x and 5x

If 8 is added to each other

3x+8:5x+8=2:3

3x+85x+8=23

3(3x+8)=2(5x+8)

9x+24=10x+16

10x-9x=24-16

x=8

Thus the numbers are 3x=3(8)=24

5x=5(8)=40

Directly using the formula, when a value is increased by R% and then decreased by R%, then net there is ( R∧2)/100 decrease. Putting R = 10, we get 1% decrease.

18/3 + 104/13 = 14

Cost price of bed= Rs.2400

Let's take market price of bed = X

As the question

C.P= (X-20X)/100

C.P= 80*X/100 -> 2400=8*X/10

X = Rs.3000.

-> S.P = 3000 ? (16*3000)/100

Selling Price = 2520

-> Profit = 2520- 2400 ==> 120Rs.

%profit = 120*100/2400= 5%

As per question, the average age of five girls is 15, Lets X be the total age of the girls

X/5= 11

X= 55

Oldest girl age+ remaining 4 girls age= 55

Remaining 4 girls age= 55-15==> 40

Average age of 4 girls = 40/4= 10 years.

Volume = πr2h

π*4*4*hA / π*5*5*hB = 12/25

Solving, hA/ hB = 3:4

Let take current age of Ram= x, current age of father = 3x

After 10 years, Ram age = x+10, Father's age= 3x+10

As per condition:

3x+10= 2*(x+10)

3x+10 = 2x+20

X= 10 years

Two numbers are in the ratio=2:7

Sum of the numbers=810

We have,

Sum of the terms in the ratio=2+7=9

First number=29×810

=2×90

=180

Second number=79×810

=7×90

=630

C.P = 1265*100*100*100/110/115/125

C.P = 800

series in the form of 5*1+1^3=6,

6*2+2^3=20,

20*3+3^3=87,

87*4+4^3=412,

412*5+5^3=125.

x:y=3:5

xy=35

5x=3y

x=3y5

3x+4y:8x+5y=3×3y5+4y:8×3y5+5y=9y+20y5:24y+25y5=29y5:49y5=29y:49y=29:49

Let's the number of pages typed in one hour by P, Q and R be p, q and r respectively. Then,

P,Q and R typed page in 1 hrs = 216/4

=> p + q + r = 216/4

=> p + q + r = 54 ...(i)

r - q = q - p => 2p = q + r ...(ii)

5r = 7p => p = 5/7 r ...(iii)

By Solving above (i), (ii) and (iii) equations

=> p = 15, q = 18, q = 21

Let the required numbers be 7x and 11x

If 7 is added to each of the numbers it becomes

7x+711x+7=23

21x+21=22x+14

X=21-14=7

Thus

The numbers are 7x=7×7=49

11x=11×7=77

Since the year is not leap year hence odd day will be = 1 day

So 1's day of next year will be = (Monday+ odd day) => Tuesday

So last day of that year = Monday

Suppose man's usual speed = s, and usual time = t

So distance = speed *time ==> d= s*t

If he travels with 5/6th of his usual speed which is = s*5/6

And time= (t+15) and distance, d = (s*5/6)*(t+15) {as distance will be same}

s*t = (s*5/6)*(t+15)

t= (5t+75)/6

t= 75 min.

We have

Sum of the terms of the ratio=2+3=5

Ravish money=25×1350

=2×270

=Rs.540

Shikha money=35×1350

=3×270

=Rs.810

(483*483*483+ 517*517*517) / (517*517 - 517*483 + 483*483) a3+b3

= (a+b)(a2-ab+b2) Hence [(a+b)(a2-ab+b2)] / (a2-ab+b2)

= a+b In the given equation a=483 and b= 517

==> 483+517=1000.

X% of SP = 34% of CP

Also, P = 26% of SP

SP - CP = 0.26(SP)

CP = 0.74(SP)

Now, (34/100)×74

X = 25.16

A & B one day work = 1/8

A alone one day work = 1/12

B alone one day work = (1/8 - 1/12) = ( 3/24 - 2/24)

=> B one day work = 1/24

so B can complete the work in 24 days.

Let ‘x’ be the monthly salary, then

(65/100 × 40/100)x + 35/100x = 43920

Solving, X= 72000

We have

Sum of the terms of the ratio=2+3+5=10

P-share= 210×totalmoney=210×2000=2×200=Rs.400

Q-share= 310×totalmoney=310×2000=3×200=Rs.600

R-share= 510×totalmoney=510×2000=5×200=Rs.1000

Given that

Three numbers are in ratio 2:3:5

Sum of these numbers=800

Sum of the terms of the ratio=2+3+5=10

Firstnumber=210×800=160Secondnumber=310×800=240Thirdnumber=510×800=400

Let the required ages be 5x and 7x

18 years ago their age ratios

5x−187x−18=813

65x−13×18=8×7x−8×18

65x-234=56x-144

65x-56x=234-144

9x=90

x=10

Thus the ages are 5x=5×10=50years

7x=7×10=70 years

x:y=8:9

xy=89

9x=8y

x=8y9

7x−4y:3x+2y=7×8y9−4y:3×8y9+2y=56y−36y9:42y9=20:42=10:21

For one year the capital is

A = (40500 * 6 + 45000 * 6) = Rs. 5,13,000

B = (45000 * 12) = Rs. 5,40,000

C = (60000 * 6 + 45000 * 6) = Rs. 6,30,000

Ratio A : B : C = 513 : 540 : 630

C's share will exceed that of A by

= [(630 - 513)/1683] * 56100 => (117/1683) * 56100

=> (117/3) * 100 = 39 * 100 = Rs. 3900

Total cost price of 40 shirts including trportation = 3000+ (3000*10)/100 = 3300

Given profit = 20%

Hence, Selling price = (100+20)*3300/100

S.P of 40 shirts = 3960

So S.P. of 1 shirt = 3960/40= Rs.99

35% of 30=25% of X+1

35*30/100= 25*X/100+1

105/10=X/4 +1

X=38

LEADING word can be written as "L" "E" "A" "D" "I" "N" "G"

Taking vowels together = E A I, so vowels can be arranged as= !3 ways and lets Vowels be a unit X

So X L D N G, It can be arranged in = !5 Ways

So complete word can be written in= !3 *! 5 = 720 ways

(A+B)'s 1-day work, when working together= 1/30

(A+B)'s 20 days' work= 20/30==> 2/3

Remaining work = (1-2/3) ===> 1/3

Remaining work was done by A in 20 more days = 1/3

So A can complete the whole work alone in = 20* 3 ===>60 days

Hence A's 1-day work, working alone = 1/60

B?s 1 day work, working alone = 1/30-1/60======1/60

So B will take 60 days to for completing the entire word.