Total hours spend by Ritika for studies =11\(\frac { 1 }{ 3 } \) hours =\(\frac { 34 }{ 3 } \)hours
 Total hours devoted by Ritika for Hindi and Sanskrit 5\(\frac { 3 }{ 5 } \)= \(\frac { 28 }{ 5 } \)hours
Total hours devoted by Ritika for other subjects
= \(\frac { 34 }{ 3 } -\frac { 28 }{ 5 } \)
= \(\frac { 170-84 }{ 15 } =\frac { 86 }{ 15 } \)
=5\(\frac { 11 }{ 15 } \) hours

(a) PQ + PA > QA
Yes, because sum of two sides of a triangle is always greater than the third side.
(b) PQ + PA > OQ + OA
Yes,because: PQ + PA > QA
PQ + PA >QO + OA [\(\Box \) QA = QO + OA]
(c) PQ + PA +AR > OQ + OA +AR
Yes, because: PQ + PA > OQ + OA
Adding AR in both sides, we get
PQ + PA + AR > OQ + OA + AR
(d) PQ + PR > OQ + OR
Yes, because
PQ + PA > QO + OA                 ...(1)
OA+AR > OR                            ...(2)
Adding (1) and (2), we get
PQ + PA + OA + AR > QO + OA + OR
PQ + PR > OQ + OR
(e) PQ + QR + PR > OP + OQ + OR
Yes, because
PQ + PR > OQ + OR          ...(1)
PQ + QR > OP + OR          ...(2)
PR + PQ > OP + OQ          ...(3)
Adding (1), (2) and (3), wet get
2(PQ + PR + QR) > 2(OP + OQ + OR).

We can write 4 x (- 8) as
4 x (- 8) = (- 8) + (- 8) + (- 8) + (- 8)
It can be represented on the number line as under:

We have, 8 x (- 2) = (- 2) + (- 2) + (- 2) + (-2) + (-2) + (-2) + (-2) + (-2)
It can be represented on the number line as under:

We have, 3 x (-7) = (-7) + (-7) + (-7)
It can be represented on the number line as under:

We have,
10 x (-1) = (-1) + (-1) + (-1) + (-1) + (-1) + (-1) + (-1) + (-1) + (-1) + (-1)
It can be represented on the number line as under:
.

106.5runs

As per the given information,
Profit on Monday = Rs 47
Loss on Tuesday = Rs 12
Loss on Wednesday = Rs 8
\(\therefore\) Net profit = Total profit - Total loss
Now, total profit = 47
and total loss =12 + 8 = 20
\(\therefore\) Net profit = 47 - 20 = Rs 27

For a pair of integers whose product is -36 and whose difference = 15.
So, first integer = -3 and second integer = 12
Their product = (-3) \(\times\) 12 = -(3 \(\times\) 12)= -36
and the difference between these two integer is 15.

Since, points A, O and B are collinear.
\(\therefore\)  (\(\angle\) AOD + \(\angle\)COD) and \(\angle\)BOC form a linear pair.
So, \(\angle\)AOD+ \(\angle\)COD+ \(\angle\)BOC=180°
\(\Rightarrow\) x-10° + 4x - 25° + x + 5° =180°
\(\Rightarrow\) x + 4x + x-10° - 25° + 5° =180°
\(\Rightarrow\) 6x-35°+5°=180°
\(\Rightarrow\) 6x - 30° = 180°
\(\Rightarrow\) 6x=180°+30° \(\Rightarrow\)6x=210°
\(\Rightarrow\) x = \(\frac { 210° }{ 6 } \) = 35°
So, \(\angle\)BOC = x + 5° = 35° + 5° = 40°
\(\therefore\)\(\angle\)BOC= 40°

Yes, given ABCD be a quadrilateral and its diagonal AC and BD intersect each other at O.

We know that, the sum of lengths of any two sides of a triangle is greater than the length of third sides.
In \(\triangle AOB\), OA + OB > AB..............(i)
Similarly in \(\triangle BOC\),OB + OC > BC...........(ii)
In \(\triangle COD\),OC + OD> CD.............(iii)
and in \(\triangle DOA\), OD + OA > DA..............(iv)
On adding Eqs. (i), (ii), (iii) and (iv), we get
(OA + OB) + (OB + OC) + (OC + OD) + (OD + OA)> AB + BC + CD + DA
\(\Rightarrow\) 2(OA + OB+ OC + OD)> AB+ BC + CD + DA
\(\Rightarrow\) AB + BC + CD + DA< 2(OA +OB + OC + OD)
\(\Rightarrow\) AB + BC + CD + DA < 2 [(OA + OC) + (OB + OD)]
\(\Rightarrow\) AB + BC + CD + DA < 2 (AC + BD)
[∵ AC = OA + OC and BD = OB + OD]

Given, total number of questions = 10
Marks for a correct answer = 5
Marks for an incorrect answer = (- 2)
Marks for not attempted the question = 0
(i) Mohan's score = (Correct questions \(\times\) Marks) + (Incorrect questions \(\times\) Marks)
= 4 \(\times\) (5) + 6 \(\times\) (- 2)
= 20 + (-12) = 20 -12 = 8
(ii) Reshmas score = 5 \(\times\) 5 + 5 \(\times\) (- 2)
=25 + (-10) = 15
(iii) Heena's score = 2 \(\times\) 5 + 5 \(\times\) (- 2) + 3 \(\times\) 0
=10 + (-10) + 0 = 0

(a) Possible situations to happen are as follows:
(i) On tossing a coin, getting either a head or a tail
(ii) On drawing one card from a pack of 52 cards one side will appear.
(iii) Getting a number from 1 to 6 by throwing a die.
(b) Impossible to happen are as follows:
(i) A girl in the boy's school.
(ii) Getting a number 8 by throwing a die.
(iii) A person of height 3 m.
(c) Mayor may not happen situations are as follows:
(i) To toss a coin and get tail.
(ii) Probably it may rain.
(iii) An ant rising to 4 m height