Tag: IGate Aptitude

Question 1. If 35% Of 30 = 25% Of X + 1, Find The Value Of X ?

Answer :

35% of 30=25% of X+1 

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

105/10=X/4 +1

X=38

Question 2. Shiva Purchased 40 Shirts For Rs.3000. He Spends 10% On Transportation. What Should Be The Selling Price Per Shirt To Gain A Profit Of 20% ?

Answer :

Total cost price of 40 shirts including transportation = 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

Question 3. If A Man Travels With 5/6th Of His Usual Speed, He Is Late By 15 Mins. What Is The Usual Time Taken To Travel ?

Answer :

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.

Question 4. If The Current Age Of Ram’s Father Is Thrice Of Age Of Ram. Then After 10 Years, The Age Of Father Will Be 2 Times The Age Of Ram. The Current Age Of Ram Is ?

Answer :

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

Question 5. The Average Age Of Five Girls In A Hostel Is 11. The Oldest Girl Among Them Is 15 Years Old. What Is The Average Age Of The Other Girls ?

Answer :

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.

Question 6. The Cost Price Of A Bed Is Rs.2400 Which Is 20% Below The Market Price If It Is Sold At A Discount Of 16% On The Market Price Then Find Its Market Price, Selling Price And Profit ?

Answer :

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%

Question 7. In How Many Ways Letters Of World “leading” Can Be Arranged, So That Vowels Always Come Together ?

Answer :

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

Question 8. If The First Day Of Year Is Monday. Then What Is The Last Day Of The Year, If It’s Not A Leap Year.

Answer :

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

Question 9. A And B Can Complete A Work In 30 Days, Working Together. Both Worked For 20 Days And Then B Left The Work. The Remaining Work Was Completed By A Alone In 20 More Days. So In How Many Days B Alone Can Complete The Entire Work.

Answer :

(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.

Question 10. Find The Value Of (483*483*483+ 517*517*517) / (517*517 – 517*483 + 483*483) ?

Answer :

(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.

Question 11. A’ And ‘b’ Complete A Work Together In 8 Days.if ‘a’ Alone Can Do It In 12 Days.then How Many Day ‘b’ Will Take To Complete The Work?

Answer :

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.

Question 12. The Respective Ratio Of Radii Of Two Right Circular Cylinders (a And B) Is 4 : 5. The Respective Ratio Of Volume Of Cylinders A And B Is 12:25. What Is The Respective Ratio Of The Heights Of Cylinders A And B?

Answer :

Volume = πr²h

π*4*4*hA / π*5*5*hB = 12/25

Solving, hA/ hB = 3:4

Question 13. The Tax On A Commodity Is Diminished By 10 % And Its Consumption Increased By 10 %. The Effect On The Revenue Derived From It Changes By K %. Find The Value Of K.

Answer :

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.

Question 14. What Percent Of Selling Price Would Be 34 % Of Cost Price If Gross Profit Is 26 % Of The Selling Price ?

Answer :

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

Question 15. P, Q And R Are Three Typists Who Working Simultaneously Can Type 216 Pages In 4 Hours. In One Hour, R Can Type As Many Pages More Than Q As Q Can Type More Than P. During A Period Of Five Hours, R Can Type As Many Pages As P Can During Seven Hours. How Many Pages Does Each Of Them Type Per Hour ?

Answer :

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

Question 16. The Retail Price Of A Water Geyser Is Rs. 1,265. If The Manufacturer Gains 10 %, The Wholesale Dealer Gains 15 % And The Retailer Gains 25 %, Then The Cost Of The Product Is:

Answer :

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

C.P = 800

Question 17. Find The Number That Can Be Put In Place Of The Question Mark 5,6,?,87,412,2185.

Answer :

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.

Question 18. A, B And C Started A Business By Investing Rs. 40,500, Rs. 45,000 And Rs. 60,000 Respectively. After 6 Months C Withdrew Rs. 15,000 While A Invested Rs. 45,000 More. In Annual Profit Of Rs. 56,100, The Share Of C Will Exceed That Of A By ?

Answer :

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

Question 19. 17.995/3.01 + 104.001/12.999 = ?

Answer :

18/3 + 104/13 = 14

Question 20. Dhruva Gave 35% Of Her Monthly Salary To Her Mother. From The Remaining Salary, She Paid 18% Towards Rent And 42% She Kept Aside For Her Monthly Expenses. The Remaining Amount She Kept In Bank Account. The Sum Of The Amount She Kept In Bank And That She Gave To Her Mother Was Rs. 43,920. What Was Her Monthly Salary ?

Answer :

Let ‘x’ be the monthly salary, then

(65/100 × 40/100)x + 35/100x = 43920

Solving, X= 72000

Question 21. If X: Y=3:5, Find The Ratio 3x+4y:8x+5y ?

Answer :

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

Question 22. If X: Y=8:9, Find The Ratio (7x-4y):3x+2y ?

Answer :

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

Question 23. Two Numbers Are In The Ratio 3:5. If 8 Is Added To Each Number, The Ratio Becomes 2:3. Find The Numbers ?

Answer :

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

Question 24. What Should Be Added To Each Term Of The Ratio 7: 13 So That The Ratio Becomes 2: 3 ?

Answer :

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

Question 25. Three Numbers Are In The Ratio 2: 3: 5 And The Sum Of These Numbers Is 800. Find The Numbers ?

Answer :

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

Question 26. The Ages Of Two Persons Are In The Ratio 5: 7. Eighteen Years Ago Their Ages Were In The Ratio 8:13. Find Their Present Ages ?

Answer :

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

Question 27. Two Numbers Are In The Ratio 7:11. If 7 Is Added To Each Of The Numbers, The Ratio Becomes 2 . 3. Find The Numbers ?

Answer :

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

Question 28. Two Numbers Are In The Ratio 2: 7.11the Sum Of The Numbers Is 810. Find The Numbers ?

Answer :

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

Question 29. Divide Rs 1350 Between Ravish And Shikha In The Ratio 2: 3.

Answer :

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

Question 30. Divide Rs 2000 Among P, Q, R In The Ratio 2:3:5

Answer :

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