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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Divide Rs 2000 Among P, Q, R In The Ratio 2:3:5 |
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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 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 |
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| 2. |
Divide Rs 1350 Between Ravish And Shikha In The Ratio 2: 3. |
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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 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 |
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| 3. |
Two Numbers Are In The Ratio 2: 7.11the Sum Of The Numbers Is 810. Find The Numbers ? |
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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 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 |
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| 4. |
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 ? |
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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 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 |
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| 5. |
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 ? |
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Answer» Let the required AGES be 5x and 7X 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 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 |
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| 6. |
Three Numbers Are In The Ratio 2: 3: 5 And The Sum Of These Numbers Is 800. Find The Numbers ? |
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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 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 |
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| 7. |
What Should Be Added To Each Term Of The Ratio 7: 13 So That The Ratio Becomes 2: 3 ? |
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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 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 |
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| 8. |
Two Numbers Are In The Ratio 3:5. If 8 Is Added To Each Number, The Ratio Becomes 2:3. Find The Numbers ? |
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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 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 |
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| 9. |
If X: Y=8:9, Find The Ratio (7x-4y):3x+2y ? |
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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 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 |
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| 10. |
If X: Y=3:5, Find The Ratio 3x+4y:8x+5y ? |
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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 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 |
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| 11. |
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 ? |
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Answer» LET ‘X’ be the MONTHLY SALARY, then (65/100 × 40/100)x + 35/100x = 43920 SOLVING, X= 72000 Let ‘x’ be the monthly salary, then (65/100 × 40/100)x + 35/100x = 43920 Solving, X= 72000 |
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| 12. |
17.995/3.01 + 104.001/12.999 = ? |
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Answer» 18/3 + 104/13 = 14 18/3 + 104/13 = 14 |
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| 13. |
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 ? |
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Answer»
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 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 |
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| 14. |
Find The Number That Can Be Put In Place Of The Question Mark 5,6,?,87,412,2185. |
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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. 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. |
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| 15. |
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: |
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Answer» C.P = 1265*100*100*100/110/115/125 C.P = 800 C.P = 1265*100*100*100/110/115/125 C.P = 800 |
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| 16. |
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 ? |
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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 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 |
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| 17. |
What Percent Of Selling Price Would Be 34 % Of Cost Price If Gross Profit Is 26 % Of The Selling Price ? |
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Answer»
ALSO, P = 26% of SP SP - CP = 0.26(SP) CP = 0.74(SP) Now, (34/100)×74 X = 25.16 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 |
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| 18. |
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. |
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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. 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. |
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| 19. |
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? |
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Answer» VOLUME = πr2h Solving, hA/ hB = 3:4 Volume = πr2h π*4*4*hA / π*5*5*hB = 12/25 Solving, hA/ hB = 3:4 |
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| 20. |
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? |
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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. 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. |
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| 21. |
Find The Value Of (483*483*483+ 517*517*517) / (517*517 - 517*483 + 483*483) ? |
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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. (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. |
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| 22. |
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. |
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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. (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. |
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| 23. |
If The First Day Of Year Is Monday. Then What Is The Last Day Of The Year, If It's Not A Leap Year. |
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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 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 |
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| 24. |
In How Many Ways Letters Of World "leading" Can Be Arranged, So That Vowels Always Come Together ? |
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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 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 |
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| 25. |
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 ? |
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Answer»
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% 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% |
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| 26. |
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 ? |
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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. 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. |
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| 27. |
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 ? |
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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 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 |
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| 28. |
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 ? |
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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. 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. |
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| 29. |
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% ? |
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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 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 |
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