12 pencils

Next two consecutive integers are x + 1 and x + 2
They are taken in increasing order and multiplied by 2, 3, and 4 respectively.
(x) × (2) = 2x
(x + 1) × 3 = 3x + 3
(x + 2) × 4 = 4x + 8
These 3 integers add up to 74,
⇒ 2x + 3x + 3 + 4x + 8 = 74
⇒ 9x + 11 = 74
⇒ 9x = 74 - 11
⇒ 9x = 63
⇒ x = 63/9 = 7
⇒ x + 1 = 8, x + 2 = 9
Thus, the three integeres are 7, 8 and 9

 From over head tank, water leaks out at 10 l/hr.
∴ Let us see how much leakage happens with time 
1 hr        10 ltrs.
2 hrs       20 ltrs.
3 hrs       30 ltrs.
4 hrs       40 ltrs.
and so on 150 min = 120 min + 30 min
= 2 hr 30 min = 2 1/2 hrs 
Water wasted in 150 min (2 1/2 hrs) = 25 litres
Time at which 75 litre of water is wasted = 7 1/2 hrs = 450 minutes.

(1) x = -7

(ii) y = 6

From the graph to form a trapezium, we have to join the points A (5, 1), B (1.5,4.5), C (- 1,2) and D (1,0).

From the graph, the points (- 2, 4) and (- 3, 5) lie on the line, which is passing through (4, - 2) and (0, 2) in the second quadrant.

a) (3, 2)
b) (- 2, 0)
c) (- 2, - 2) 
d) (- 2, 2)
e) (-1, -1)
f) (3, -1)
g) (0, 2)
h) (2, 0)
i) (- 3, 3) 
j) (0, -2)

\(\frac { 13 }{ 21 } \)

l = 8 cm, b = 24 cm

3, 21

Rs.8000

700cm

Let ABCD be a rhombus. The diagonals of the rhombus meet at O.
Since \(\triangle\) AOB is a right angled triangle
AB² = AO² + OB² 
= 82 + 62
= 64 + 36
= 100
\(\mathrm{AB}=\sqrt{100}=10\)
​​​​​​The perimeter of the rhombus = 4 x AB
= 4 x 10
= 40cm

From the figure AC is to be found.
By using Pythagoras theorem,
AC² = AB² + BC² = 282 + 212 = 784 + 441 = 1225 = 352
∴ AC = 35 km

From the figure
d² = 80² + 150²
= 6400 + 22500
d² = 28900
d = 170
The distance between the helicopter and the ship is 170 m.

25, 24

(i) From the figure,
x² = 9² + 40²
= 81 + 1600
= 1681
\(x=\sqrt{1681}\)
x = 41
(ii) From the figure
34² = 30² + y²
1156 = 900 + y²
y = 1156 - 900
y² = 256
\(y=\sqrt{256}\)
y = 16
(iii) From the figure
39² = 36² + z²
1521 = 1296 + Z²
z² = 1521- 1296
z² = 225
\(z=\sqrt{225}\)
z = 15

i) School bag:
M.P = Rs. 500
Discount = 5 %
S..P = M.P - 57% of M.P
\(=500-\frac{5}{100} \times 500\)
S.P = 500 - 25
S.P = 475
GST \(=\frac{12}{100} \times 475=RS. 57\)
ii) Hair dryer:
M.P = Rs. 2000
Discount = 10%
S.P = 2000 - 10% of 2000
\(=2000-\frac{10}{100} \times 2000\)
= 2000 - 200
= 1800
GST = 28% of 1800
\(=\frac{28}{100} \times 1800\)
= Rs. 504

Total bill amount = Rs. 2836

We know that if the,cost price of x articles = selling price of y articles, then
\( \text { Profit } \% =\left(\frac{x-y}{y} \times 100\right) \% \)
\(=\left(\frac{15-10}{10} \times 100\right) \% \)
\(=\frac{5}{10} \times 100 \%=50 \%\)

Let x be the cost price of the article.
810 - x = x - 530
2x = 810 + 530
2x = 1340
x = 670
The cost price of the article is Rs. 670.

Let the number be x.
Given x + 18% of x = 236
\( x+\frac{18}{100} x =236 \)
\(x\left(1+\frac{18}{100}\right) =236 \)
\(x\left(\frac{100+18}{100}\right) =236 \)
\(x\left(\frac{118}{100}\right) =236 \)
\(\therefore x =\frac{236 \times 100}{118} \)
= 200
The required number is 200.

Let the numberbe x.
Given 75% of x - 60% of x = 82.5
\( \frac{75}{100} x-\frac{60}{100} x =82.5 \)
\(\frac{x}{100}(75-60) =82.5 \)
\(\frac{x \times 15}{100} =82.5 \)
\(\therefore x=\frac{82.5 \times 100}{15}\)
x = 550
The required number is 550
\(20 \% \text { of } 550=\frac{20}{100} \times 550\)
= 110

(i) 500
ii) Cricket -45,Volleyball-42, Badminton -33

Let the number be x.
\( \frac{32}{100} \times x =48 \)
\(x =\frac{48 \times 100}{32}=150\)
The number is 150