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

     


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