Important for :

1

A

Let the road from Kashmir Valley to Leh be downhill for a distance of x km, and uphill for a distance of (S – x) km. (Where S = distance between the two towns).

Now, on the return journey (S – x) km will be downhill and x km will be uphill.

Hence total distance downhill = x + (S – x) = S

And total distance uphill = (S – x) + x = S

Hence, total time = | S | + | S | = 9 hour (Given) |

40 | 80 |

⇒ (2 × S + S) = (80 × 9)

⇒ 3 × S = (80 × 9)

⇒ S = | ( | 80 × | 9 | ) | = 240 |

3 |

Thus, the distance between Kashmir Valley and Leh = 240 km

Hence, option (A) is correct.

2

C

The interest I on principle P for D days at R% p.a. is given as
I = P × D × | 2R |

73000 |

From 1st January to 9th January,

P = Rs.20000, D = 9 and R = 4% p.a.

Therefore, interest, I1 = | 20000 × 9 × 8 | = | 1440000 | = Rs. 19.73 |

73000 | 73000 |

From 10th January to 9th February,

P = Rs.19000, D = 31 and R = 4% p.a.

Therefore, interest, I2 = | 19000 × 31 × 8 | = | 4712000 | = Rs. 64.55 |

73000 | 73000 |

From 10th February to 9th March,

P = Rs.18000, D = 28 and R 4% p.a.

Therefore, interest, I3 = | 18000 × 28 × 8 | = | 4032000 | = Rs. 55.23 |

73000 | 73000 |

From 10th March to 9th April,

P = Rs.17000, D = 31 and R = 4%p.a

Therefore, interest, I4 = | 17000 × 31 × 8 | = | 4216000 | = Rs. 57.75 |

73000 | 73000 |

From 10th April to 9th May,

P = Rs.16000, D = 30 and R = 4% P.a.

Therefore, interest, I5 = | 16000 × 30 × 8 | = | 3840000 | = Rs. 52.60 |

73000 | 73000 |

From 10th May to 6th June,

P = Rs.15000, D = 27 and R = 4% p.a.

The days are 27 as on the day of closing he will not get the interest. So 22 days of May and 5 days of June will be counted.

Therefore, interest, I6 = | 15000 × 27 × 8 | = | 3240000 | = Rs. 44.38 |

73000 | 73000 |

Total Interest- = I1 + I2 + I3 + I4 + I5 + I6 = 19.73 + 64.55 + 55.23+ 57.75 + 52.60 + 44.38

= Rs.294.24 = Rs.294 approximately

Hence, option (C) is correct.

3

D

Let money borrowed by Aman is x.If Rohan didn’t apply compound interest, amount payable by Aman after 7 years is.

∴ x + |
x × 7 × 10 | = | 17x |

100 | 10 |

Now Rohan applied SI for first 4 years and for remaining 3 years CI is applicable, then amount payable by Aman is.

For first 4 years.

∴ x + |
x × 4 × 10 | = | 14x |

100 | 10 |

For next 3 year CI is payable

∴ |
14x | ( | 1 + | 10 | ) |
^{3} |
= | 18634x |

10 | 100 | 10000 |

Now Aman has to pay 3268 more.

∴ |
18634x | – | 17x | = 3268 |

10000 | 10 |

∴ |
1634x | = 3268 |

10000 |

Hence, option D is correct.

4

B

Let Jethalal make n knots. Tarak Mehta makes (n – 10) knots. Since both the Dhagas start and end with a knot, the length of the Dhaga L given by

L = 10(n – 1) =12 (n – 11)

⇒ n = 61 and L = 600

Hence, option (B) is correct.

Hence, option (B) is correct.

5

A

Let the number or notes of denominations Rs.50, Rs.20 and Rs.10

with Prem Kumar be x, y and z respectively.

Total money with Prem Kumar = (200 – 30) = 170

(50x + 20y +10z) = 170.

Since x, y, z > 1, x cannot be 3 therefore x = 2.

When x = 2, y can be only 2 otherwise if y = 3, then x Will be equal to 1, which is not possible.

Therefore x = 2, y = 2 and z = 3.

The total number of notes with Prem Kumar is 7.

Alternative Solution:

We can directly start with two notes of each denomination, which gives us 100 + 40 + 20 = Rs.160. Now, the only possibility of

having exactly Rs.170 is to have just one additional note of Rs.10.

having exactly Rs.170 is to have just one additional note of Rs.10.

Hence, 7 notes.

Therefore, option (A) is correct.

Therefore, option (A) is correct.