IMO - Mock Test - 5

The series is as follows:

13 × 7 = 91
(13 + 4) × 7 = 17 × 7 = 119
(17 + 4) × 7 = 21 × 7 = 147
(21 + 4) × 7 = 25 × 7 = 175
(25 + 4) × 7 = 29 × 7 = 203

COLD = OCDL
Here, the positions of the first and the second letters get interchanged with each other and also the positions of the third and the fourth letters are interchanged.

So, KEPT = EKTP

'how' means '+', 'you' means '-', 'are' means '÷', and 'today' means '×'.

Then, 20 how 10 are 5 you 4 today 2 = 20 + 10 ÷ 5 - 4 × 2

Using BODMAS,

20 + (10 ÷ 5) - 4 × 2 = 20 + 2 - (4 × 2) = 20 + 2 - 8 = 14

When the letters of the given word ENCOURAGE are arranged in reverse alphabetical order, then the result will be:

URONGEECA

So, the letter that is eighth from the left is C.

The series is as follows:
2² = 4
3³ = 27
4² = 16
5³ = 125
6² = 36
Hence, ? = 7³ = 343

He was ranked 5^th from the top and 40^th from the bottom.
So, total number of students = 5 + 40 – 1
= 45 – 1 = 44

First, we dig the soil and put the seed in the soil, so 1 comes first.
Secondly, plant grows, so 3 comes second.
Thirdly, plant becomes tree, so 5 comes third.
Then, from tree, we get wood and after that we make table from wood.
That means, the sequence becomes: 1, 3, 5, 2, 4.

48 - 6 + 8 x 2 ÷ 10 = ?
Replacing the values accordingly, we get
48 ÷ 6 x 8 - 2 + 10
Using BODMAS,
8 x 8 - 2 + 10
= 64 - 2 + 10
= 62 + 10 = 72

Given: n = 73, a = 9 and a₇₃ = 297
a_n = a + (n - 1)d
So, 9 + (73 - 1)d = 297
9 + 72d = 297
72d = 288
d = 288/72
d = 4
a₄₅ = a + 44d = 9 + 44(4) = 9 + 176 = 185

The numbers of flowers in the flower bed rows form an AP.
a = 12
d = 2
Let the number of rows be n.
So, 40 = 12 + (n – 1) × 2
14 = n – 1
Or n = 15

Solving the equation:

x² + 4x + 3 = 0

x² + 3x + x + 3 = 0

(x + 1)(x + 3) = 0

We get

x = -1, -3

Sum of the roots (S) = (-1) + (-3) = -4

 

Mean of 5 observations = 7
Total of 5 observations = 7 x 5 = 35
Mean of 6 observations = 7
Total of 6 observations = 7 x 6 = 42
Therefore, sixth observation = 42 - 32 = 7

According to the question,
(x + 16)² = 2x² + 368
Or x² + 256 + 32x = 2x² + 368
Or x² - 32x + 112 = 0
Or (x - 28)(x - 4) = 0
Or x = 28 or 4

Let a = 6, b = 8, c = 12
Then, a² = 6² = 36
b² = 8² = 64
c² = 12² = 144

Now, a² + b² = 100
So, a² + b² < c²

Hence, it is an obtuse-angled triangle.

(x + 7)² – 6(x + 7) + 9

Put x + 7 = a ......... (1)

a² – 6a + 9 = (a – 3)²

Put the value of a from (1).

= [(x + 7) – 3]² = [x + 7 – 3]² = (x + 4)² = (x + 4)(x + 4)

x² – kx + 20 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -k and c = 20

The roots will be imaginary if the discriminant b² – 4ac < 0.

k² – 80 < 0

k² < 80

The greatest integer k which satisfies this is 8.

Since, angles are in ratio 6 : 9, measure of the first angle will be 6x.
Then, measure of the second angle will be 9x.
According to the question
6x + 9x = 90^o
15x = 90^o
x = 90^o/15
x = 6^o
Hence, measure of the first angle = 6x = 6 × 6 = 36^o
Measure of the second angle = 9x = 9 × 6 = 54^o
So, the largest angle = 54^o

Given that x - 3y = 8 ------------------ (1)

2x - 5y = 4 ------------ (2)

From (1),

x = 8 + 3y

Putting this value in (2), we get

2(8 + 3y) - 5y = 4

16 + 6y - 5y = 4

y = -12

Now, x = 8 + 3(-12)

So, x = 8 - 36 = -28

Possible values of degree are: 0, 1, 2, 3 or 4.
So, sum of possible values of degrees of p(x) and q(x) = 10.

As x - 1 is a factor of p(x) = 4x³ + 3x² - 4x + k
So, p(1) = 0
So, p(1) = 4(1)³ + 3(1)² - 4(1) + k = 0
4 + 3 - 4 + k = 0
k = -3

Initially, Deepika had 27 coins.
Then, she finds p coins from the drawer, that means, she now has 27 + p coins.
Priyanka has double the number of coins Deepika has.
That means, she has 2 x (27 + p) = 54 + 2p
Altogether, they have (27 + p) + (54 + 2p) = 81 + 3p

Let the number of elephants = Number of men = x

Then, number of legs = 4x + 2 x (x/2) = 5x
So, 5x = 80, or x = 16

Cost price of the article = Rs. 5,000
Selling Price of the article = Rs. 1.5 × 5,000 = Rs. 7,500
Selling Price of the article after a month = Rs. 0.7 × 7,500 = Rs. 5,250
Hence, the price after discount = Rs. 5,250.

Total ball pens = 100
Defective ball pens = 10
Pens which are good = 100 - 10 = 90
Then, probability of drawing a good pen = 90/100 = 9/10

Current time = 11:15 am
Time of last flight = 11:15 - 50 minutes = 10:25 am
So, time of next flight = 10:25 am + 6 hours = 4:25 pm

Write the percentage as a decimal
35% = 0.35
The mark-up price is:
488 × 0.35 = 170.80
The new price will be
170.80 + 488.00 = 658.80
Hence, the new price after mark-up will be Rs. 658.80.

Let the cost of one pen = Rs. x
cost of one pencil = Rs. y
So, according to the question
8x + 10y = 90 ----- (1)
7x + 8y = 78 ------- (2)
So, from equations (1) and (2)
we get, x = 10 and y = 1
So, the cost of one pen is Rs. 10.

Let CP of the mixer be Rs. x.
Concept to be used: SP – CP = Profit
And, CP – SP = Loss

Given: When SP = Rs. 56, a profit is made.
Then, Profit = Rs. (SP – CP) = Rs. (56 – x)

When SP = Rs. 42, loss is incurred.
Loss = CP – SP = Rs. (x – 42)

Since, profit made = loss incurred

So, 56 - x = x - 42
56 + 42 = 2x
98 = 2x
x = 49

So, CP of the mixer = Rs. 49

number of pins is 20. Hence
X + Y = 20.... (1)
If X is the number of pins that cost $0.25, then the cost of X pins will be 0.25 X.
If Y is the number of pins at $0.15, then the cost of Y pins will be 0.15 Y.
The total cost of X pins and Y pins is known to be $4.20, and is also given by
0.25 X + 0.15 Y = 4.2 ....(2)
Solve (1) and (2), we get
X + Y = 20
0.25 X + 0.15 Y = 4.2
The first equation gives Y = 20 - X.
Substitute value of (Y) in the second equation and solve
0.25 X + 0.15 (20 - X) = 4.2
X(0.25 - 0.15) + 3 = 4.2
0.1 X = 1.2
X = 12 and Y = 20 - 12 = 8

If we arrange the data in ascending order:- 34.5, 36.5, 36.9, 38.6, 39.0, 40.5, 40.6, 42.8
Median = (38.6+39/2) = 38.8 kg
Thus, statement I is not true.
Range of data = 42.8 – 34.5 = 8.3 kg
statement II is true.

1. An arc is a semi- circle when its ends are the ends of a diameter.
2. Segment of a circle is the region between an arc and chord of the circle.
3. A circle divides the plane, on which it lies, in three parts - exterior, interior and circumference.