Mixed Quantitative Aptitude Questions Set 135

1. A father divides his property between his two sons A and B. A invests his amount at C.I. of 8% per annum whereas B invests the amount at S.I. at 10% per annum. At the end of 2 years, the interest received by B is Rs. 1336 more than the interest received by A. Find the share of A in the father’s property of Rs. 25,000.
   Rs 12,000
   Rs 14,000
   Rs 15,000
   Rs 10,000
   Rs 16,000
   Option D
   Let share of A = Rs x
   Share of B = Rs (25000 – x)
   [(25000 − 𝑥)×10×2]/ 100 − 𝑥 [(1 + 8/ 100)^2 − 1] = 1336
   ⇒ (25000 − 𝑥)/ 5 − 104𝑥/ 625 = 1336
   ⇒ 3125000 – 125x – 104x = 1336 × 625
   ⇒ x = Rs 10,000
   
   
    



2. An ore contains 25% of an alloy that has 90% iron. Other than this, in the remaining 75% of the ore, there is no iron. To obtain 60 kg of pure iron, the quantity of the ore needed, in kgs, is approximately:
   232.31 kg
   214.25 kg
   266.67 kg
   242.34 kg
   256.15 kg
   Option C
   Required quantity of ore = 100/ 25 × 100 /90 ×60 = 266.67 kg
   
    



3. The strength of a school increases and decreases in every alternate year by 10%. It started in 2000 and increases in 2001, then the strength of the school in 2003 as compared to that in 2000 was:
   Increase by 7.1%
   Increase by 6.5%
   Decrease by 7.8%
   Increase by 7.6%
   Increase by 8.9%
   Option E
   Let in 2000, the strength was 100
   In 2001 strength = 110
   In 2002 strength = 110 × 90/ 100
   In 2003 strength = 110 × 90/ 100 × 110 /100 = 108.9
   Required % increment = 8.9%
   Hence, strength after 3 years will increase by 8.9%
   
   
    



4. A spider climbed 62(1/2)% of the height of the pole in one hour and in the next hour it covered 12(1/2)% of the remaining height. If pole’s height is 192 m, then the distance climed in second hour is:
   7
   9
   5
   6
   8
   Option B
   Total height = 192m
   Distance climbed in second hour = 1 /8
   = 192 × (8−5)/ 8 × 1 /8 = 192 × 3 /8 × 1/ 8 = 9 m
   
   
    



5. C is twice efficient as A. B takes thrice as many days as C. A takes 12 days to finish the work alone. If they work in pairs (i.e. AB, BC, CA) starting with AB on the first day then BC on the second day and AC on the third day and so on, then how many days are required to finish the work?
   4 (1/8) days
   2 (1/7) days
   3 (1/5) days
   5 (1/9) days
   6 (1/4) days
   Option D
   Time taken by A alone = 12 days
   Time taken by B alone = 12 2 ×3 = 18 days
   Time taken by C alone = 12 2 = 6 days
   Work done in first three days = ( 1 /12 + 1 /18) + ( 1/ 18 + 1/ 6 ) + ( 1/ 6 + 1/ 12)
   = 5 /36 + 4 /18 + 3 /12 = 11/ 18
   Remaining work = 1 – 11/ 18 = 7/ 18
   Work done in next two days = 5/ 36 + 4/ 18 = 13/ 36
   Now, remaining work = 7 /18 – 13/ 36 = 1 /36
   This work will be done by C and A together
   Total days required = 3 + 2 + 4/ 36 = 5 (1/9) days
   
   
    



6. 3, 52, 88, 113, 129, ?
   144
   124
   138
   161
   155
   Option C
   3, 52, 88, 113, 129, ?
   +7^ 2 , +6^ 2 , +5^ 2 , +4^ 2 , +3^ 2
   ? = 138
   
   
    



7. 2, 3, 8, ?, 112, 565
   20
   15
   27
   22
   17
   Option C
   2, 3, 8, ?, 112, 565
   × 1 + 1,× 2 + 2,× 3 + 3,× 4 + 4,× 5 + 5
   ? = 27
   
   
    



8. 6, 4, 8, 23, ?, 385.25
   77.7
   81.4
   84.5
   90.6
   78.5
   Option C
   6, 4, 8, 23, ?, 385.25
   × 0.5 + 1,× 1.5 + 2,× 2.5 + 3,× 3.5 + 4,× 4.5 + 5
   ? = 84.5
   
   
    



9. 8, 64, 216, 512, ?, 1728
   1400
   1500
   1300
   1000
   1600
   Option D
   8, 64, 216, 512, ?, 1728
   2^ 3 , 4^ 3 , 6^ 3 , 8^ 3 , 10^3 , 12^3
   ? = 1000
   
   
    



10. 1, 1, 2, 6, 24, 120, 720, ?
    3553
    4141
    5040
    5500
    4884
    Option C
    1, 1, 2, 6, 24, 120, 720, ?
    × 1,× 2,× 3,× 4,× 5
    ? = 5040